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Ministry of Higher Education & Scientific Research
University of Baghdad – College of Engineering
Mechanical Engineering Department
Incompressible Fluid Mechanics
Prof. Dr. Ihsan Y. Hussain
Mech. Engr. Dept. / College of Engr. – University of Baghdad
February 2017 Jamada AlAoula 1438
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 2
Preface
The present book is a handout lectures for the B.Sc. Course
ME202 : Fluid Mechanics / I. The course is designed for B.Sc. Sophomore Students in
the Mechanical Engineering Discipline. The time schedule needed to cover the course
material is 32 weeks, 3 hrs. per week. The course had been taught by the author (course
tutor) for more than 25 years. A short c.v. for the author is given below;
Prof. Dr. Ihsan Y. Hussain / Baghdad  1964
B.Sc. ( 1986 ), M.Sc. ( 1989 ), & Ph.D. ( 1997 ) in Mechanical
Engineering from the Mech. Engr. Dept. – University of Baghdad
Professor of Mechanical Engineering – ThermoFluids
Teaching Undergraduate Courses and Laboratories in Various
Iraqi Universities ( Baghdad, AlKufa, Babylon, Al  Nahrain ….)
in Various Subjects of Mechanical Engineering
Teaching Advanced Graduate Courses (M.Sc. and Ph.D.) in various
Iraqi Universities (Baghdad, Technology, Babylon, AlKufa, AlMustansyrya, Al
Nahrain…) in the Areas of ( Fluid Mechanics, Heat Transfer, CFD, Porous Media, Gas
Dynamics, Viscous Flow, FEM, BEM
Lines of Research Covers the Following Fields ;
Aerodynamics
Convection Heat Transfer ( Forced, Free, and Mixed )
Porous Media ( Flow and Heat Transfer )
Electronic Equipment Cooling
Heat Transfer in Manufacturing Processes ( Welding, Rolling, … etc. )
Inverse Conduction
Turbomachinery ( Pumps, Turbines, and Compressors )
Heat Exchangers
Jet Engines
PhaseChange Heat Transfer
Boundary Layers ( Hydrodynamic and Thermal )
Head of the Mech. Engr. Dept. / College of Engineering  University of Baghdad ( December
/ 2007 – October / 2011 )
Member of Iraqi Engineering Union (Official No. 45836).
ASHRAE Member ( 8161964 )
Head of ( Quality Improvement Council of Engineering Education in Iraq QICEEI )
Supervised ( 41 ) M.Sc. Thesis and ( 20 ) Ph.D. Dissertations
Publication of more than (70) Papers in the Various Fields Mentioned above
Member in the Evaluation and Examining Committees of more than ( 300 ) M.Sc. and Ph.D.
Students in their Theses and Dissertations
Member of the Editing Committee of a number of Scientific Journals
Evaluation of more than ( 700 ) Papers for Various Journals and Conferences
Working within the Mechanical Groups in the Consulting Engineering Bureau, College of
Engineering, University of Baghdad, and in the Dept. of Mech. Engr. , in the Mechanical
Design of Various Projects
Contact Information :
Email: [email protected] : [email protected] [email protected]
Skype Name : drihsan11
Mobile No. : +9647901781035 : +9647705236582
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 3
"Table of Contents"
Section Page
No.
Table of Contents 3
I. Preface 7
I.1 General Approach 7
I.2 Learning Tools 7
I.3 Contents 9
I.4 Time Scheduling Table 10
I.5 Laboratory Experiments 10
I.6 Textbook and References 11
I.7 Assessments 11
Chapter 1 : Introductory Concepts 12
1.1 Definition of Fluids and Fluid Mechanics 12
1.2 Scope and Applications of Fluid Mechanics 14
1.3 Dimensions and Units 15
1.4 Newton's Law of Viscosity for Fluids 16
1.5 Basic Definitions and Concepts 18
Examples 25
Problems 27
Chapter  2 : Fluid Statics 28
2.1Introduction 28
2.2 Pressure Variation in a Static Fluid 28
2.2.1 Pressure Variation with Direction 28
2.2.2 Pressure Variation in Space 29
2.2.3 Basic Hydrostatic Equation 30
2.2.4 Pressure Variation for Incompressible Fluids 31
2.2.5 Pressure Variation for Compressible Fluids 31
2.3 Pressure Measurements 32
2.3.1 Absolute and Gage Pressure 32
2.3.2 Atmospheric Pressure Measurements 33
2.3.3 Gage Pressure Measurements 34
2.3.3.1 Bourdon Gage 34
2.3.3.2 Pressure Transducers 34
2.3.3.3 Manometers 34
2.3.3.3.1 Piezometer 35
2.3.3.3.2 Differential Manometers 36
2.3.3.3.3 Micro Manometer 36
2.3.3.3.4 Inclined Manometer 36
2.4 Hydrostatic Forces on Submerged Surfaces 37
2.4.1 Horizontal Surfaces 37
2.4.2 Inclined Surfaces 38
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 4
2.4.3 Curved Surfaces 40
2.5 Buoyancy 42
2.5.1 Hydrometers 44
2.6 Stability of Submerged and Floating Bodies 44
2.6.1 Submerged Bodies 44
2.6.2 Floating Bodies 45
2.6.3 Metacenter 46
2.7 Relative Equilibrium 47
2.7.1 Uniform Linear Acceleration 47
2.7.2 Uniform Rotation 49
Examples 50
Problems 55
Chapter 3 : Fluid Flow Concepts 56
3.1 Definitions and Concepts 56
3.2 General Control Volume Equation 61
3.3 Law of Conservation of Mass: Continuity Equation 63
3.3.1 Continuity Equation at a Point 63
3.4 Law of Conservation of Energy: Energy Equation 64
3.4.1 Bernoulli's Equation 67
3.5 Law of Conservation of Liner Momentum: Momentum Equation 69
3.5.1 Applications of Momentum Equation 70
3.5.1.1 Fixed Vanes (Blades) 70
3.5.1.2 Moving Vanes (Blades) 71
3.5.1.2.1 Single Moving Vane 71
3.5.1.2.2 Series of Moving Vanes 72
3.5.1.3 Forces on pipe Bends 73
3.5.1.4 Theory of Propellers 74
3.5.1.5 Jet Propulsion 75
3.5.1.6 Rocket Mechanics 76
3.5.2 Euler's Equation of Motion 77
3.5.3 Bernoulli's Equation 77
3.6 Some Applications of Continuity, Momentum and Energy Equations 78
3.6.1 Losses due to Sudden Expansion 78
3.6.2 Hydraulic Jump 79
3.7Law of Conservation of Angular Momentum: Moment of Momentum
Equation
79
Examples 81
Problems 87
Chapter 4 : Dimensional Analysis And Similitude 88
4.1 Introduction 88
4.1.1 Dimensions and Units 88
4.2 Dimensional Analysis 90
4.2.1 The Buckingham II  Theorem 91
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 5
4.2.2 Common Dimensionless Numbers 93
4.3 Similitude 96
4.3.1 Re  Criterion 98
4.3.2 Fr  Criterion 98
4.4 Ships Models Tests 99
Examples 101
Problems 103
Chapter 5 : Viscous Fluid Flow 104
5.1 Introduction 104
5.2 Equations of Motion for Viscous Flow 104
5.3 Laminar Flow Between Parallel Plates: Couette Flow 106
5.4 Losses in Laminar Flow 108
5.5 Laminar Flow Through Circular Tubes and Annuli 110
5.5.1 Circular Tubes : Hagen  Poiseuille Equation 111
5.5.2 Annuli 112
5.6 Boundary  Layer Flow 113
5.6.1 Laminar Boundary  Layer 114
5.6.2 Turbulent Boundary  Layer 115
5.6.3 Boundary  Layer Separation 116
5.7 Drag an Lift 118
5.8 Resistance to Flow in Open and Closed Conduits 119
5.8.1 Open Channels Flow 121
5.8.2 Steady Incompressible Flow Through Pipes 121
5.8.3 Simple Pip Problems 123
5.8.4 Secondary (Minor) Losses 124
Examples 127
Problems 135
Chapter 6 : Fluid and Flow Measurements 136
6.1 Density Measurements 136
6.2 Pressure Measurements 136
6.2.1 Static Pressure measurements 137
6.2.2 Total (Stagnation) Pressure Measurements 138
6.3 Velocity Measurements 138
6.4 Discharge Measurements 140
6.4.1 Orifice Meters 140
6.4.1.1 Orifice in Reservoirs 140
6.4.1.2 Orifice in Pipes 143
6.4.2 Venturi Meter 144
6.4.3 Rotameter 145
6.5 Viscosity Measurements 145
6.5.1 Rotating Concentric Cylinders 145
6.5.2 Capillary Flow method 146
6.5.3 Say bolt Viscometer 146
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 6
Examples 148
Problems 150
Chapter 7 : Closed Conduits Flow Networks 151
7.1 Introduction 151
7.2 Pipe Friction Formula 151
7.3 Concept of Hydraulic and Energy Grade Lines 152
7.4 Combination of Pipes 154
7.4.1 Pipes in Series 154
7.4.2 Pipes in Parallel 155
7.4.3 Branching Pipes 156
7.4.4 Pipe Networks 157
7.5 Pumping Stations 158
7.5.1 Single Pump in Pipe Line 158
7.5.2 Pumps in Series 160
7.5.3 Pumps in Parallel 161
7.6 Conduits with non  Circular Cross  Section 162
Examples 163
Problems 171
Appendix –A : Course Folio 172
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 7
I. Preface
I.1 General Approach
The present “Lecture Notes" are intended for use in a first course in “Fluid
Mechanics" for under  graduate engineering student in the sophomore years. It represents
the revision of lecture notes for about 25 years experience in teaching the subject of
(Fluid Mechanics / I) in the second year at the Mechanical Engineering Department /
College of Engineering  University of Baghdad and other Iraqi universities. The text
covers the basic principles of fluid mechanics with a broad range of engineering
applications.
The students are assumed to have completed the subject of (Engineering Mechanics
/ Statics & Dynamics) and (Mathematics / I) courses. The relevant concepts from theses
two subjects are introduced and reviewed as needed. The emphasis throughout the text is
kept on the "physics" and the “physical arguments" in order to develop an “intuitive
understanding" of the subject matter.
Throughout the text, we tried to match between engineering education and
engineering practice. The text covers all the standard topics in fluid mechanics with an
emphasis on physical mechanisms and practical applications, while de – emphasizing
heavy mathematical aspects, which are being left to computers. We try to encourage
“Creative Thinking" and development of a “Deeper Understanding" of the subject matter
by the students.
I.2 Learning Tools
The emphasis in the text is on "developing a sense of underlying physical
mechanism" and a “mastery of solving practical problems" an engineer is likely to face in
the real world. Thus, the text covers more material on the fundamentals and applications
of fluid mechanics. This makes fluid mechanics a more pleasant and worthwhile
experience for the student.
The principles of fluid mechanics are based on our “everyday experiences" and "
experimental observations". A more physical intuitive approach is used through the text.
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 8
Frequently, “Parallels are drawn" between the subject matter and students' every day
experiences so that they can relate the subject matter to what they already know.
The material in the text is introduced at a level that an average student can follow
comfortably. It speaks "to" the students, not "over" the students. All derivations in the
text are based on physical arguments with simple mathematics, and thus they are easy to
follow and understand.
Figures are important learning tools that help the students “get the picture". The text
makes effective use of graphics.
A number of worked –out examples are included in the chapters of the text. These
examples clarify the material and illustrate the use of basic principles. An intuitive and
systematic approach is used in the solution of the example problems.
A number of sheets of solved problems that covers the material of the subject are also
included in the appendices. These problems are focusing on the application of concepts
and principles covered in the text in solving engineering problems related to important
application of the subject.
At the end of each chapter, a number of selected unsolved problems from related
chapters in the “textbook" used during the course (reference (1)) in the list of references)
are listed. The problems are grouped under specific topics (articles) in the order they are
covered to make problem selection easier for the student.
A collection of examination paper is also included in the appendices. The collection
covers quizzes, midterm exams, comprehensive exam and final exams. It serves as a tool
to improve students’ way of thinking and to make them familiar with the nature of
examinations questions.
A number of quizzes are usually made during the teaching period of the subject (about
32 weeks). These many quizzes serve the following;
1. Make these students familiar with the environment of the examinations.
2. Continuous study and followup from students to the material of the subject.
3. A “bank" of miscellaneous questions covering the various applications of the subject
material will be available for both students and instructor.
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 9
4. The final average mark will be distributed on large number of quizzes and tests, so
that, the shortage in one or more of these quizzes will not affect the evaluation greatly.
A number of laboratory experiments (710) are made to cover the principle and
concepts of the subject, experimentally. These experiments help to give the student skills
of dealing with the devices set up, acquiring and recording data, writing reports, analysis
and discussion of the results, conclusions and recommendations.
I.3 Contents
The lecture note comprises seven chapters and a number of appendices. The first five
chapters cover the basic fundamentals of the material, while the last two ones involve the
important applications of the subject.
In chapter 1 we introduce the basic fundamental concepts and definitions of fluid
mechanics as a science and its applications in the real world. Chapter 2 presents the
basic concepts of fluid statics and its application in engineering and industry. It includes
the derivation of hydrostatic pressure variation, pressure measurements, forces on
submerged surfaces, buoyancy and stability of immersed and floating bodies and relative
equilibrium. In chapter3, the fundamentals of fluid motion are studied. The governing
continuity, energy and momentum equations are derived by using the Reynoldstransport
theorem. The various applications of these equations are introduced. Chapter4 covers the
dimensional analysis and similitude principles and their applications. In chapter5, the
real (viscous) fluid flow is considered. Various applications of this type of fluid flow are
investigated, which include: laminar flow between parallel plates and through circular
tubes and annuli, boundarylayer flow, drag and lift, Moody diagram and simple pipe
problems. Chapter6 concerns the measurements of fluid and flow properties. These
include density, pressure, velocity, flow rate and viscosity measurements. In chapter 7,
the analysis of pipes and pumps connection is introduced. Series, parallel and branching
pipes networks are analyzed. Besides, series and parallel pumps connection in pumping
stations is considered. The appendices include solved sheets of problems, collection of
examination papers sheets of problems, collections, collection of examination papers and
test questions.
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 10
I4 Time Scheduling Table
The subject teaching period is (32 weeks), (4 hrs.) per week (3hrs. theory and 1 hr.
tutorial). The following is the time scheduling table for the subject weekly outlines, see
Table (I.1). This time schedule is for the theoretical part of the subject only.
Table (I.1): Time Scheduling Table
Week Covered Articles Tests Week Covered Articles Week
1
Ch
ap
ter
1
1.1+1.2+
1.3+1.4 17
C
hap
ter
Ch
ap
ter
5


4
4.1+4.2+4.2.1
2 1.5 Q1 18 4.2.2+4.3+4.3.1 Q11
3
Ch
ap
ter
2
2.1+2.2 Q2 19 4.3.2+4.4 Q12
4 2.3 20 5.1+5.2+5.3 Q13
5 2.4.1+2.4.2 Q3 21 5.4+5.5 Q14
6 2.4.3 22 5.6+5.6.1+5.6.2 Q15
7 2.5 Q4 23 5.6.3+5.7 Q16
8 2.6.1+2.6.2 Q5 24 5.8+5.8.1 Q17
9 2.6.3 25 5.8.2 Q18
10 2.7.1 Q6 26 5.8.3+5.8.4 Q19
11 2.7.2 27
Ch
ap
ter
6
6.1+6.2+6.3 Q20
12
Ch
ap
ter
3
3.1+3.2+3.3 Q7 28 6.4 Q21
13 3.4 29 6.5 Q22
14 3.5+3.5.1.1
to 3.5.1.3 Q8 30
Ch
ap
ter
7
7.1+7.2+7.3 Q23
15
3.5.1.4 to
3.5.1.6+
3.5.2+3.5.3
Q9 31 7.4 Q24
16 3.6+3.7
Q10
+
TEST
I
32 7.5+7.6
Q25
+
Test
II
Note: The experimental part of the subject (1 hr. per week) in addition to
the 4 hrs. (3 theory + 1 tutorial), see the following article.
I.5 Laboratory Experiments
We should mention here that the experimental part of the subject (1 hr per week) is
included in a separate general “Mechanical Engineering Laboratories/II" course. It
involves a number of experiments covering the principles and concepts of the subject
material. As was mentioned earlier, the experimental part aims to acquire the student a
skill of dealing with the devices, recording test data, writing a report, analysis and
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 11
discussion of the results, presentation of the experimental observations and results, and
conclusions and recommendations drawn from the results. The following is a list of some
of these experiments;
1. Dynamic Similarity.
2. Meta centric Height.
3. Impact of Jet.
4. Friction in Pipes.
5. Minor Losses.
6. Flow Measurements (Orifice and Venturi Meters).
7. Stability of Floating Bodies.
8. Bernoulli’s Equation Demonstration.
I.6 Textbook and References
The text adopted in teaching the subject is;
"Fluid Mechanics", by victor L.Streeter and E. Benjamin Wylie; First SI metric Edition,
Mc.Graw Hill, 1988
Other references which may be used are listed below;
1. "Elementary Fluid Mechanics", by John k.Vennard and Robert L. Street, 5th
edition, John Wylie and Sons,1976
2. "Engineering Fluid Mechanics", by John A. Roberson and Clayton T. Crow, 2nd
Edition, Houghton Mifflin Co.,1998
I.7 Assessments
The final mark (100%) is distributed along the teaching period as follows;
Activity Mark
Quizzes (1520 Nos.) 15%
Comprehensive Tests (2Nos.) 10%
Extracurricular Activities 5%
Final Test 70%
Total Sum 100%
The Passed Average Required is (50%)
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 12
Chapter 1
Introductory Concepts
1.1 Definition of Fluids and Fluid Mechanics
According to the variation of volume and shape with pressure, the state of matter is
usually classified in to three states, solid, liquid and gas, see Fig. (1). Fluids in clued the
liquids and gases.
Fluids
Fig. (1)
Fluid
It is a substance that deforms continuously when subjected to shear stress, no
matter how small that shear stress may be. Fluids may be either liquids or gases. Solids,
as compared to fluids, cannot be deformed permanently (plastic deformation) unless a
certain value of shear stress (called the yield stress) is exerted on it.
According to the variation of density of the fluids with pressure, fluids are
classified in to "incompressible" and "compressible" fluids.
Solid
Fixed Volume
Fixed Shape
Molecular Spacing ≃ 1 °A
(1 °A =1010
m)
Liquid
Fixed Volume
Variable Shape
Molecular Spacing ≃ 102–103 °A
Gas
Variable Volume
Variable Shape
Molecular Spacing ≃ 104 °A
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 13
Incompressible Fluids
They are the fluids with constant density, or the change of density with pressure is
so small that can be neglected and considers the density as constant. The incompressible
fluids are basically the "LIQUIDS". Gases at low velocities are usually considered as
incompressible fluids also.
There are no exact incompressible fluids in practice. For example, the density of
water at atmospheric pressure (0.1 MPa) is (1000 kg/m3). When the pressure is increased
to (20 MPa), the density becomes (1010 kg/m3). Thus, increasing the pressure by a factor
of (200) increases the density by only (1%)!! For this reason, it is reasonable to consider
the liquids as incompressible fluids with constant density.
Compressible Fluids:
They are the fluids with variable density, or the change of density with pressure is
large and cannot be neglected. These include basically the "GASES". In some liquids
problems, such as "water hammer", the compressibility of liquids must be considered.
Fluid Mechanics
It is the science that deals with the action of forces on fluids, which may be either
liquids or gases. Fluid mechanics, as a science, is a branch of the "Engineering
Mechanics" as a general science deals with the action of forces on bodies or matter
(solids, fluids (liquids of gases)), see Fig. (2).
Fig. (2)
Engineering Mechanics
Mechanics of Rigid
Bodies
Solids
Kinematics Kinetics
Mechanics of Deformable
Bodies
(Strength of Material)
Mechanics of Fluids
(Liquids & Gases)
Fluids
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 14
1.2 Scope and Applications of Fluid Mechanics
Fluids and Fluid Mechanics play a vital role in our daily life. Large and diverse
applications which are based on fluid mechanics exist. Typical examples are listed below;
1 Irrigation.
2 Navigation.
3 Power Generation (Hydraulic, Gas and Steam Power Plant).
4 Ships, Boats and Submarines.
5 Airplanes and Hovercrafts:
a. Wing Surfaces to Produce Lift.
b. Jet Engines to Produce Thrust.
c. Fuselage Design for Minimum Drag.
d. Various Systems in the Air Craft (A/c, Fuel, Oil, Pneumatic).
e. Control of the Airplane (Tail, Flaps, Ailerons, …).
6 Cars and Motorcycles.
a. Pneumatic tires.
b. Hydraulic Shock Absorbers.
c. Fuel System (Gasoline + Air).
d. Air Resistance Grates Drag on Car.
e. Lubrication System.
f. Cooling System.
g. Aerodynamic Design of Car Profile for Minimum Drag.
7 Design of Pipe Networks.
8 Transport of Fluids.
9 Air – Conditioning and Refrigeration Systems.
10 Lubrication Systems.
11 Design of Fluid Machinery (Fans, Blowers, Pumps, Compressors, Turbines,
Windmills, ….).
12 Bioengineering (Flow of Blood through Veins and Arteries).
13 Fluid Control Systems.
14 All Living Creatures Need Water (Fluid) for Life (We Made from Water Every
Living Thing).
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 15
1.3 Dimensions and Units
"Dimensions" are physical variables that specify the behavior and the nature of a
certain system, whereas the "Units" are used to specify the amount of these dimensions.
Generally, two systems of units exist. These are the International System of units (SI) and
the United States of units (U.S.). The base quantities in the (SI) system are the Mass (M),
length (L) and time (T), therefore it is called the (MLT system), whereas in the (U.S.)
system, the base quantities are the force (F), Length (L) and time (T), and is called the
FLT system. The dimensions are expressed by capital letters. Table (1) lists the
dimensions and units of the most important variables in Fluid Mechanics, in both (SI) and
(U.S.) systems, followed by conversion factors between the two systems.
Table (1): Dimensions and Units
Physical
Quantity Dimension
Units
SI U.S.
Mass M kg Slug
Length L m ft
Time T s sec.
Force F N lbf
Temperature o C
K
o F (Ordinary
Temp.) o R (Absolute
Temp.)
Area L2 m
2 ft
2
Velocity LT1
m/s ft/sec. (fps)
Acceleration LT2
m/s2 ft/sec
2
Work FL
(ML2T
2)
N.m (J) lbf . ft
Power FLT
1
ML2T
3
N.m /s (
), W lbf.ft / sec.
Pressure FL
2
ML1
T2
N/m
2 (Pa) lbf / ft
2 (psf)
Density ML3
kg / m3 Slug / ft
3
Conversion Factors
1 Slug = 14.59 kg
Slug= 32.2 lbm
1 lbm = 0.4536 kg
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 16
2 1 ft = 0.3048 m
1 ft = 12 in
1 in = 2.54 cm
1 mile = 5280 ft
1 mile = 1609 m.
3 1 lbf = 4.448 N
1 N = (1 kg) (1 m/s2)
1 lbf = (1 Slug) (1 ft/sec2)
4 K= oC + 273
oR=
oF + 460
oF = 1.8
oC + 32
oR=1.8 K
Prefixes
G = 109 gega m = 10
3 milli
M = 106 mega = 10
6 micro
k = 103 kilo n = 10
9 nano
c = 102
centi p = 1012
pico
1.4 Newton's Law of Viscosity for Fluids
Consider a fluid placed between two closely spaced parallel plates so large that
conditions at their edges may be neglected. The lower plate is fixed and a force (F) is
applied at the upper plate which exerts a shear stress (F/A) on any fluid between the
plates, see Fig. (3). The force (F) causes the upper plate to move with a steady velocity
(U), no matter how small the magnitude of (F)
Fig. (3)
The fluid in the immediate contact with a solid boundary has the same velocity as the
boundary; i.e., there is no slip at the boundary. The fluid in the area (abcd) flows to the
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 17
new position (ab'c'd), each fluid particle moving parallel to the plate at the velocity (u)
varies uniformly from zero at the stationary plate to (U) at the upper plate. Experiments
show that, other quantities being held constant, (F) id directly proportional to (A) and (U)
and is inversely proportional to thickness (h). In equation form;
Thus;
…………………...... (1.1)
Where; = Proportionality constant, viscosity of the fluid
A = Surface area parallel to the force F
Defining the shear stress ( = F/A), then;
...... (1.2)
The ratio (U/h) is the angular velocity of line (ab), or it is the rate of angular deformation
of the fluid, i.e., the rate at which the angle (bad) decreases. The angular velocity may be
written as (du/dy) as;
(
)
(du / dy) is more general and may be visualized as the rate at which one layer moves
relative to an adjacent layer. Thus, equ. (1.2) may be written as;
(Newton's Law of Viscosity) .................. (1.3)
Materials other than fluids, such as solids and plastics, cannot satisfy the definition of
fluid and equ. (1.3), since they do not deform continuously unless their "yield shear
stress" is applied.
Fluids may be classified as "Newtonian" or ” Non  Newtonian". In Newtonian
fluid, there is a linear relation between the applied shear stress ( ) and the resulting
angular deformation (du/dy) (i.e., is constant in equ. (1.3), see Fig. (4). In Non 
Newtonian fluids, the relation is non  liner. An “ideal plastic" has a definite yield stress
Incompressible Fluid Mechanics 2017
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and a constant linear relation of ( ) to (du/dy). A “thixotropic" substance, such as
printer's ink, has a viscosity that is dependent upon the immediate prior angular
deformation of the substance and has a tendency to take a set when at rest. Gases and thin
liquids tend to be Newtonian fluids, while thick, long  chained hydrocarbons may be non
 Newtonian fluids.
Yield Stress
Thixotropic
substance
Fig. (4): Rheological Diagram
1.5 Basic Definitions and Concepts
1 Density( )( )
It is the mass (m) of a fluid per unit volume ( );
…………....... (1.4)
For water at standard pressure (760 mmHg) and 4 °C :
For air at standard pressure (760 mmHg) and 20 °C :
2 Specific Volume ( )
It is the volume occupied by unit mass of fluid; i.e., it is the reciprocal of the
density ( );
……………...... (1.5)
3 Specific Weight (γ) (gama)
It is the force of gravity (weight) (W) of a fluid per unit volume ( ). It is also
called the unit gravity force.
γ
γ = g
,
………….......(1.6)
𝜇 𝜇0 (𝑑𝑢
𝑑𝑦)
(𝜇0 𝑛) 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠
𝜏
𝑑𝑢
𝑑𝑦
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(γ) Changes with location depending upon gravity (g).
γ water =9806 N/m3 at 5 °C at sea level
γ air = 11.8 N/m3 at 20 °C and patmstd = 101325 Pa
4 Specific Gravity (s)
It is the ratio of specific weight of a substance to that of water; it is also the ratio of
the mass of a substance to the mass of an equivalent volume of water at standard
condition. It may also be expressed as a ratio of its density to that of water.
s =
(dimensionless) ……………...... (1.7)
The specific gravity is also called the "relative density”. In equ. (1.7), ( ) is taken as
( N/m3) at standard reference temperature (4 °C). For gases, air or oxygen
are usually taken as the references fluid instead of water.
5 Pressure (p)
It is the normal force (F) pushing against a plane area divided by the area (A), see
Fig. (5);
(
( )
( )) ....
(1.8)
If a fluid exerts a pressure against the walls of a container, the container will exert
a reaction on the fluid which will be compressive. Liquids can sustain very High
compressive pressure, but they are very week in tension. The pressure intensity is a scalar
quantity, has a magnitude only. Pressure may be express in terms of an equivalent height
(h) of fluid column;
p = γ h ……………………… (1.9)
6 Shear Stress ( ) (taw)
It is the shear force (F) acts upon the surface (parallel) area (A), see Fig. (6)
( ) (
) …………...... (1.10)
Fig. (5)
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Unlike pressure, shear stress may have two components, see Fig. (6);
…………....... (1.10a) (z xy  plane)
………………..... (1.10b)
7. Viscosity
It is the property of a fluid by virtue of which it offers resistance to shear. There
are two types of viscosity; "Dynamic (or Absolute) viscosity ( ) , and the "Kinematic
Viscosity". Their definitions are;
......................... (1.11) (from Newton's law of viscosity, equ. (1.3))
......................... (1.12)
Units of Viscosity
SI Units:
( )
Poise P =
(
)
U.S. Units:
(
)
Causes of Viscosity:
1. Cohesion forces between molecules, (effective in liquids).
2. Exchange of momentum by molecules in the layers, (effective in gases).
Notes:
1. Viscosity of gases increases with temperature, but viscosity of liquids decreases with
temperature, see Fig. (C.1) and (C.2) p.p (536537) in your textbook.
Fig. (6)
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2. For ordinary pressure, viscosity is independent of pressure.
3. Kinematic viscosity represents the momentum diffusion coefficient; its value for gases
is larger than that for liquids.
8 Specific Heat (c)
It is the property that decries the capacity of a substance to store thermal energy. It
is the amount of heat that must be transmitted to a unit mass of a substance to raise its
temperature by one degree.
For gases, (c) depends on the process accompanied the change in temperature;
accordingly, there are two types;
Specific heat at constant pressure.
Specific heat at constant volume.
Specifics' heat ratio ........………....... (1.13)
R = Gas constant .......................... (1.14)
Examples: Water:
Air:
9 Vapor Pressure ( )
It is the pressure at which a liquid will boil. Vapor pressure is a function of
temperature (it increases with temperature). When the liquid pressure is less than the
Fig. (c.1)
p.p536 in textbook
Fig. (c.2)
p.p.537 in text book
𝜇 𝑣
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vapor pressure, liquid flashes into vapor. This is the phenomenon of "cavitation" which
results in the “erosion" of the metal parts.
Ex. Water at (20 °C) has = 2.447 kPa
At (100 °C) has = 101.3 kPa
Mercury at (20 °C) has = 0.173 kPa
10 Equation of State: Perfect Gas
The perfect gas is defined as a substance that satisfies the “perfect gas law"; which has
the following forms;
p =mRT …………………… (1.15a)
p =RT …………………… (1.15 b)
p= RT …………………… (1.15c)
p =nRT …………………… (1.15d)
Where;
R=
…………………………….. (1.15e)
And =
……………………… (1.15f)
p= Absolute Pressure (Pa)
T=Absolute lute Temperature (K)
R= Gas constant (J/kgK)
R=Universal Gas Constant =8312 J/mol K
Mw=Molecular Weight (kg/ mol)
n=Number of Moles
Real gases below critical pressure and above critical temperature tend to obey the
perfect gas law. The perfect law encompasses both Charles' law (which states that for
constant pressure ( ) and Boyle's law (which states that for constant
temperature : )
11 Bulk Modulus of Elasticity (k)
It expresses the compressibility of a fluid (liquids and gasses). It represents the
change of volume (d ) caused by change in pressure (dp). It is defined as;
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k =
(Pa) …………………………………… (1.16)
Where:
dp=Change in pressure (Pa)=p2p1
d =Change in volume (m3) =
d =Charge in density (
) =
=Original volume (m3) =
=Original density (kg/m3) =
For perfect gases;
P= RT
=RT K=
= RT K=p
Ex. Kwater=2.2 Gpa
K air=0.0001 Gpa
12 Surface Tension ( ) (sigma)
At the interface between a liquid and a gas, or two immiscible liquids, a film or
special layer seems to form on the liquid, apparently owing to attraction of liquid
molecules below the surface. It is a simple experiment to place a small needle on a water
surface and observe that it is supported there by the film.
The formation of this film may be visualized on the basis of “surface energy" or
work per unit area required to bring the molecules to the surface. The "surface tension "
is then the stretching force required to form the film, obtained by dividing the surface
energy term by unit length of the film in equilibrium. The surface tension of water varies
from about (0.074
) at (20ºC) to (0.059
) at (100ºC).
Capillarity
Capillarity attraction is caused by surface tension and by the relative value of
“adhesion"(Which is the attraction between liquid particles and solid) to "cohesion"
(which is the molecular attraction between liquid particles), see Fig. (7).
If dp>0 (p2>p1), then:
d < ( < )
d𝜌 > (𝜌 > 𝜌 )
The capillarity rise (h) is usually calculated by applying equilibrium equation to
the capillary tube shown in Fig.(8), and as follows;
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∑
cos =w
2 cos =
Thus;
h=
……………. (1.17)
Where;
h = Capillary rise (m)
Wetting angle [for clean tubes,
= ( 0o ) for water and (140º) for mercury]
Note:
1 For d>12 mm, capillary effects are negligible.
2 Equ (1.17) is plotted in Fig. (1.4) in your textbook.
Some Applications of Surface Tension
The action of surface tension is to increase the pressure within droplet, bubble and
liquid jet. To calculate the pressure sustained in these cases, a force balance is made, and
as follows;
1 Droplet
For a section of half of spherical droplet, see Fig. (9);
=
P* =2
Thus:
……………………………. (1.18)
2 Bubble
For a section of half of a spherical bubble, see Fig (10);
= +
p* =2*
Thus;
p=
………………….. (1.19)
Fig. (8)
𝐹𝜎
Fig. (9)
Fig. (10)
𝐹𝜎
𝐹𝜎
𝐹𝑝
Fig. (7)
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3 Liquid Jet
For a section of a half of a cylindrical liquid jet, see Fig. (11) ;
= +
p*L2r= =2
Thus p=
…………………. (1.20)
4 Measurements of ( ): Ring Pulled out of a Liquid
For a ring wetted by a liquid, see Fig. (12);
F=W+
F=W+ ( )………………….. (1.21)
Equ. (1.21) is usually used to measure ( ). The weight (W) is usually negelcted.
Examples
Example (1.1): The weight of the cylinder shown in the figure is (W) and its radius of
gyration is (KG). The cylinder rotates at angular speed (N). Develop an expression for
viscosity of oil ( ) required to stop the cylinder in (10) seconds.
Sol.:
T= T1+T2
T1= F*R=
∴ T1=
dT2=
∴ T2 =∫
⁄
0
∴T=
∑
∴
0
Fig. (11)
𝐹𝜎
Fig. (12)
𝐹𝜎0
𝐹𝜎
𝐹𝑝
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Example (1.2): Derive an expression for the torque (T) required to rotate the circular cone
shown in the figure at a rate of (ω), in terms of the related variables ( , ω, and R).
Sol.:
dF=
=
=
( )
dF=
dT=dF*r
=
T=∫
0
=
∴T=
Example (1.3): A reservoir of glycerin has a mass
of (1200 kg) and have a volume of (0.952 m3). Find the glycerin weight, density, specific
weight and specific gravity.
Sol.:
W=mg=1200*9.8 W=11770 N
= 00
0 =1261 kg/
=
0
0
= =1261*9.8
Example (1.4): Convert (15.14 Poise) to kinematic viscosity in ( ) if the liquid has
a specific gravity of (0.964).
Sol.:
=s =0.964*1000 =964 kg/
γ = 12360
N/𝑚
sin𝜃
𝑟 𝑠𝑖𝑛𝛽
β
β
𝜃
𝜃 𝜇
𝜇
𝑟 𝑠𝑖𝑛𝛽
𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑐𝑜𝑛𝑒
𝜃
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15.14P *
0
( )
Example (1.5): The bulk modules of elasticity of water is (2.2 GPa), what pressure is
required to reduce its volume by (0.5 percent).
Sol.:
k=
2.2*109=
0 0
0 00
Example (1.6): A small circular jet of mercury (0.1 mm) in diameter issues from an
opening. What is the pressure difference between the inside and outside of the jet? Use
( N/m)
Sol.:
∑
p*DL=2
Thus
p=
0
0 0
Problems
The problems number listed in the table below refer to the problems in the
"textbook", Chapter "1”;
Article No. Related Problems
1.4 1,2,3,5,14,15,16,17,22
1.5 7,9,11,12,28,39,42,43,46,47
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Chapter 2
Fluid Statics
2.1 Introduction
In static fluids, no relative motion between the fluids particles exist, therefore no
velocity gradients in the fluid exist, and hence no "shear stresses" exist. Only "normal
stresses (pressure)" exist. In this chapter, the pressure distribution in a static fluid and its
effects on surfaces and bodies submerged or floating in it will be investigated.
2.2 Pressure Variation in a Static Fluid
The pressure (p) in a static fluid may change with space coordinates (x, y, z) and
with direction ( ).Thus, to find the differential change in pressure (dp) in a static fluid, we
may write;
p= p (x, y, z, )
…….… (2.1)
To derive the differential equation for the pressure (p),
we have to derive an expressions For the partial
derivatives (
) in equation (2.1),
Which represents the changes in pressure with (x, y, z and ), see Fig. (1).
2.2.1 Pressure Variation with Direction
The pressure at a point in a static fluid acts with the
same magnitude in all directions. This physical fact will
be proved mathematically by taking a fluid element of
tetrahedron shape, see Fig. (2). we will apply Newton's
2nd
law in the three directions, knowing that;
Where l, m, n = direction cosines
+ ∑
Fig. (1)
𝑝𝑥 yz l ne ( re δAx)
𝑝𝑦 xz l ne ( re δAy)
𝑝𝑧 xy l ne ( re δAz)
𝑝 l ne ( re δA)
Fig. (2)
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+ ∑
+ ∑
Thus; independent of direction
If we take the limit as [( ) ], the element approaches a "point", thus, the
above result applies for a point in a static fluid, that is the pressure is independent of
direction, and hence;
………………. (2.2)
2.2.2 Pressure Variation in Space
To find the pressure variation in a static fluid with space coordinates (x, y, z), a
cubical fluid element will be studied, see Fig. (3). The pressure (p) at the center of the
element is assumed to change with the three coordinates and we want to find the rate of
changes (
). The Newton's 2
nd law is applied for the element of Fig. (3).
H.O.T (Higher Order Term)
Fig. (3)
(𝑝 𝜕𝑝
𝜕𝑦
𝑑𝑦
)𝑑𝑥 × 𝑑𝑧
(𝑝 𝜕𝑝
𝜕𝑧
𝜕𝑧
)𝑑𝑦𝑑𝑥
(𝑝 𝜕𝑝
𝜕𝑥
𝑑𝑥
𝑧)𝑑𝑦𝑑𝑧
)
d
(𝑝 𝜕𝑝
𝜕𝑦
𝑑𝑦
)𝑑𝑥𝑧 (𝑝
𝜕𝑝
𝜕𝑧)𝑑𝑦𝑑𝑥
(𝑝 𝜕𝑝
𝜕𝑥
𝑑𝑥
)𝑑𝑦𝑑𝑧
0
𝑝(𝑥 𝑦 𝑧)
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+∑
(
) (
)
Or;
Since
…………………….. (2.3)
+ ∑
(
) (
)
= 0 ………………………. (2.4)
Equations (2.3) and (2.4) indicate that no changes in pressure occur in the
horizontal xz plane, i. e., all points in the same horizontal plane in a static fluid have
the same pressures.
+ ∑
(
) (
)
Or;
………………………………………. (2.5)
Equation (2.5) indicates that the pressure in a static fluid decreases as (y), the
vertical coordinate, increases. This is a logical result physically, since the increase in (y)
means decease in the fluid column height which causes the pressure.
2.2.3 Basic Hydrostatic Equation
This equation is obtained by substituting equations (2.2), (2.3), (2.4) and (2.5) into
equation (2.1). Thus;
0
Equ. (2.3)
Equ. (2. )
γ Equ. (2.5)
0 Equ. (2.4)
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Hence: y
Or;
This is the basic differential equation whose integration gives the pressure
distribution equation in a static fluid. The integration depends on how the specific weight
( ) vary with the vertical coordinate (y). Accordingly, the distribution for
incompressible and compressible fluids is different.
2.2.4 Pressure Variation for Incompressible Fluids
For incompressible fluids, ( =constant), and hence ( = constant), and the
integration of equ. (2.6) yields;
∫ ∫ y
y n
Or; y n
The ( n ) of integration is the same
for any two points. (1) and (2), thus;
y y
y
If we use the coordinate (h = y) measured vertically downward, i. e.,
(h = y2), then see Fig (4);
…………………………… (2.7)
As mentioned previously, the pressure increases as we move downward (+h) and
decreases as we move upward (h).
2.2.5 Pressure Variation for Compressible Fluids
For compressible fluids, the density ( ) changes with temperature (T) as the perfect
gas law (p = RT) indicates. And since the temperature (T) varies with altitude (y), the
density ( ) also changes with (y), and accordingly ( ) changes with (y) also. Thus;
………………….. (2.8)
Basic Hydrostatic Equation …………………. (2.6)
0 (y1=0)
Fig. (4)
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Hence, equ. (2.6) becomes;
……………… (2.9)
To integrate this equation, we must know how (T) varies with (y). For this purpose, the
atmosphere is divided into two layers;
I  Troposphere: (from sea level up to 10769m)
0 ( 0) …………………. (2.10)
Where: 0= Reference temperature at seal eve ( 0 ) = 288k
= Lapes rate = 0.00651 ˚C/m = 0.00357 ˚F/ft
Thus, equ. (2.9) becomes;
( )
Integrate from ( )where( ), to ( ) where( ). Thus:
( )
………………. (2.11)
II  Stratosphere: (from y = 10769 m up to 32 km)
( y r e ) …….... (2.12)
Thus:
And hence:
( )
……………. (2.13)
( ) is the pressure at ( )
2.3 Pressure Measurements
2.3.1 Absolute and Gage Pressures
The absolute pressure ( ) is the pressure measured relative to a complete
vacuum (zero absolute pressure), whereas the gage pressure ( ) is the pressure
measured relative to the local atmospheric pressure ( ) ).The local atmospheric
pressure may be larger than, smaller than are equal to standard atmospheric pressure
( ) ) which is a constant value, see Fig. (5). Thus;
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) ………………… (2.14)
2.3.2 Atmospheric Pressure Measurement
The "barometer" is usually used to measure the atmospheric pressure. It consists of
a glass or Perspex tube with one open and immersed in a bath of mercury, see Fig. (6). the
pressure at the free surface is ( ), which is equal to the pressure at point 2 ( ) (see
equs. (2.3), and (2.4)). Thus, using equ. (2.7) ( );
…………………….. (2.15)
The vapor pressure of mercury is very small and can be neglected,
hence;
………………………………. (2.16)
And;
……………………….. (2.17)
If ( ) ), then (h = 76 cm). The mercury is used because its specific weight is
large and thus (h) will be small and reasonable. Also the vapor pressure is very small and
can be neglected.
"Aneroid Barometer" is also used to measure the
atmospheric pressure, see Fig. (7). It measures the
difference in pressure between the atmospheric pressure
( ) and an evacuated cylinder by means of a sensitive
elastic diaphragm and linkage system.
Fig. (5)
Fig. (6)
Fig. (7)
Pabs=0
Absolute zero
(Complete Vacuum)
𝑝𝑎𝑏𝑠 𝑝𝑎 1.01325bar
76 cmHg 10.34mH2o
1 atmosphere
14.7 psi(𝑙𝑝𝑓
𝑖𝑛 )
2116 psf (𝑙𝑝𝑓
𝐹𝑡 )
29.92 in Hg 33.91 ft H2o
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2.3.3 Gage Pressure Measurements
2.3.3.1 Bourdon Gage
The "bourdon gage" shown in Fig. (8) is a typical
of the devices used for measuring gage pressure. The
pressure element is a hollow, curved, flat metallic tube
closed at one end; the other end is connected to the
pressure to be measured. When the internal pressure is increased,
the tube tends to straighten, pulling on a linkage to which is attached a pointer and causing
the pointer to move. The dial reads zero when the inside and outside of the tube are at the
same pressure, regardless of its particular value. Usually, the gage measures pressure
relative to the pressure of the medium surrounding. The tube, which is the local
atmosphere. According to equ. (2.7); the pressure recorded by the gage of Fig. (8) is;
…………………….. (2.18)
…………………….. (2.19)
2.3.3.2 Pressure Transducers
The devices work on the principle of conversion pressure energy to mechanical
energy, then to electrical signals, sees Fig. (9).
2.3.3.3 Manometers
Manometers are devices that employ liquid columns for determining difference in
pressure. They are usually made of glass (or PVC) tubes filled of one or more liquids with
different specific gravities and that are not miscible in each other. Each manometer has
tow ends. The general procedure that should be followed in working any manometer
problem is as follows;
Fig. (8)
Fig. (9)
Fig. (8)
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1. Start at one end (or any meniscus if the circuit is continuous) and write the pressure
there in an appropriate unit or in an appropriate symbol if it is unknown.
2. Add to this the change in pressure, in the same unit, from one meniscus to the next
(plus if the next meniscus is lower, minus if higher).
3. Continue until the other end of the gage (or the starting meniscus) is reached and
equate the expression to the pressure at that point, known or unknown.
2.3.3.3.1 Piezometer
This type of the manometers, illustrated in Fig. (10),
is used to measure the pressure in a liquid when it is above zero
gage. Using equ. (2.7);
h (Pa)
= ( ) ................… (2.20)
( ) .................… (2.21)
Piezometer would not work for negative gage pressure, because air would flow into
the container through the tube. It is also impractical for measuring large pressures at A,
since the vertical tube would need to be very long.
For measurement of small negative or positive gage pressures, the tube may take
the form shown in Fig. (11). For this arrangement;
( )
= ( )
( ) (2.23)
( )( )
For greater negative or positive gage pressure, a second
liquid of greater (s) is employed, see Fig. (12).
=
Fig. (11)
Fig. (12)
……….. (2.22)
……….. (2.24)
……….. (2.25)
Fig. (10)
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Fig. (13)
2.3.3.3.2Differential Manometer
This type of manometers (Fig. (13)) determines the difference in pressure at two
points A and B. the equation of this monometer is:
Or;
( ) .… (2.26)
And;
(Units of length of ) .... (2.27)
Equation (2.26) represents the difference in both absolute and gage pressures at A and B.
2.3.3.3.3 Micromanometers
This type of manometers are used for measuring
small pressure differences, see Fig . (14). The small
difference ( ) at C.S.A.(A) is magnified to a
large difference (R/2) at smaller C.S.A.(a).
The manometer equation is;
( ) (
)
(
) ( )
But; A
Thus, we can show that:
* (
)
+ ……… (2.28)
2.3.3.3.4 Inclined Manometer
The inclined manometer (Fig. (15)) is
frequently used for measuring small difference
in gage pressure. It is adjusted to read zero, by
moving the inclined scale. Since the inclined tube
requires a greater displacement of the meniscus
for given pressure difference than a vertical tube,
𝑠 𝛾
𝑠 𝛾
𝑠 𝛾
Fig. (14)
𝑠 𝛾
𝑦
Fig. (15)
𝑠
𝛼
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it offers greater accuracy in reading the scale. The equations of the manometer are;
( )
Thus;
( )
but h=R sin
and A
Thus;
( n
) ………………. (2.29)
The device is usually designed so that the angle ( ) can be changed to more than one
value.
Note:
The value of pressure (p) can be converted to columns of any fluid by the following
relation;
p =
Thus;
Unit length of fluid 1
Unit length of fluid 2 ............………..... (2.30)
Unit length of fluid 3
2.4 Hydrostatic Forces on Submerged Surface
Hydrostatic forces excreted by a static fluid are always perpendicular to the surface
on which they act.
2.4.1 Horizontal Surfaces
A plane surface in a horizontal position in a fluid at rest is
subjected to a constant pressure ( p = ), see Fig.(16). To find the
resultant hydrostatic fore (F), consider an element (dA), on which
an element force (dF) acts;
dF = pdA = A
Thus; F = ∫ ∫ A ∫ A
Fig.(16)
∇
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Hence;
F= A A ... (2.31)
Center of Pressure C.P. ( )
It is the point through which the resultant hydrostatic force (F) passes. To find the
coordinates of center of pressure ( ), take moments about x & y coordinates;
F*x = ∫
A x ∫ ∫x A
Hence; x
∫ x x
F * y ∫
y ∫ ∫y A
Hence; y
∫ y
Hence; C.P. = C
(x y ) = (x y) [for horizontal surfaces only] …………...(2.32)
2.4.2 Inclined Surfaces
For inclined plane surfaces, the pressure (p= ) varies anlog the surface, since (h)
is not constant, see Fig. (17). For element (dA), the force acting on it (dF) is;
𝑦 n𝛼
h= y sin 𝛼 Fig. (17)
p= 𝛾
𝑝𝑐= 𝛾
If (𝛼 °)(vertical surfacse) → h=y ; =𝑦
∇ α
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dF= pdA= n
F=∫ ∫ y n n ∫ y n A y n
Hence;
F = ………….... (2.33)
To find the center of pressure (x y );
F * y = ∫
y n y ∫ n
Thus; y
∫
Where = second moment of area (A) about horizontal axis (ox).
From the parallel axis theorem;
Where Second moment of area (A) about horizontal centroidal axis (cx).
= Distance between the two axes (ox) & (cx)
Thus;
y
………………….. (2.34)
F * x = ∫
y n x ∫ n
Thus; x
∫
Where;
= Product moment of area (A) about axes (ox & oy)
= Product moment of area (A) about centroidal axes (cx & cy)
= (parallel axis theorem)
Thus;
x
............................……………….... (2.35)
When x or y or both are axes of symmetry, then ( ) & (x )
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The Pressure Prism
It is a prismatic volume with its base the given surface area and with altitude at any
point of the base given by (p = ), see Fig. (18)
dF = pdA = A
F= ∫ ∫ …………...... (2.36)
Thus, the volume of the pressure prism equals the
magnitude of the resultant force acting on one side of the surface.
To find cent of pressure;
F * xP ∫ x
∫ ∫
Similarly, we can show that;
i.e. C.P. = C of volume
( ) ……..…………... (2.37)
2.4.3 Curved Surfaces
When the elemental force (dF = pdA) vary in direction, as in curved surfaces, they
must be added as vector quantities; i.e., their components (horizontal and vertical) are
added vectorially. In Fig. (19), the force (dF) on an element (ds) of the curved surface is
replaced by its two components, horizontal (dFH) and vertical (dFV).
Fig. (19)
Horizontal Component (FH)
The horizontal component of pressure force on a curved surface is equal to the
pressure force exerted on a projection of the curved surface. The vertical plane of
projection is normal to the direction of the component
Fig. (18)
𝛾 𝐴
𝛾 𝐵
∇
γ
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dF = pdA =
dFH = dF sin n
Thus; FH= ∫ ∫ ............… (2.38)
Where;
Vertical projection of the curved surface on a vertical plane
Centroid of ( ) distance to the free surface
∫
…..… (2.39)
∫
………… (2.40)
Equations (2.38), (2.39) and (2.40) are similar to equations (2.33), (2.34) and (2.35) for
inclined surfaces, with ( °), i.e., vertical surface, which is the vertical projection
( ) of the curved surface.
Vertical Component (Fv)
The vertical component of pressure force on a curved surface is equal to the weight
of liquid vertically above the curved surface and extending up to the free surface;
d y y x
Thus; ∫ ∫ ……………… (2.41)
Where;
l e l en l e e een e r e r e n e ree r e
∫
∫
∫
i.e.;( ) ( )
( ) ( ) .........…………….……... (2.42)
Note;
When the liquid is below the curved surfaces, see Fig. (20), and the pressure
magnitude is known at some point o (p0), an "imaginary" or equivalent free surface (I.S.)
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can be constructed (
) above (o). The direction of force ( ) is reversed to obtain the
real fore (F). The resultant of ( ) and ( ) is found as;
F = √
=
......................…(2.43)
2.5 Buoyancy
The buoyant force ( ) is the resultant force exerted on a body by a static fluid in
which it is submerged or floating. Its magnitude is equal to the weight of the displaced
volume. It acts vertically upward through the center of buoyancy (B) which is the centroid
(C) of the displaced volume. To prove the above statements mathematically, consider the
body shown in Fig. (21). the body has 6 curved surfaces, on which 4 horizontal forces
(F1,F2, F3 andF4) and 2 vertical forces (F5 andF6) are acting. To find the resultant of these
forces;
∑
=
= 0 (since )
∑
=
= 0 (since )
∑

R = √
Thus : ................................ (2.44)
To find the center of buoyancy (B), take moments;
Fig. (20)
𝑝𝑜 γ
Fig. (21)
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∫
∫ x ∫
Similarly, we can show that
Hence; ( ) ( )
( ) ( ) …………………................ (2.45)
Notes:
1 When the body is submerged in two fluids, then the buoyant force ( ) is the sum of
the two buoyant forces exerted by the two fluids, see Fig. (22)
=
2 When the body is floating, only the submerged volume is used in calculating ( ), see
Fig. (23)
∑
Thus;
l l e y
l e l e
y e e
3 To find the volume ( ) and weight (W) of an oval  shaped body, it is weighted in two
fluids ( ), see Fig. (24)
W=
=
Weight of body in fluid 1
Weight of body in fluid 2
Fig. (22)
Fig. (23)
Fig. (24)
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2.5.1 Hydrometers
They are devices that use the principle of buoyant force to determine the specific
gravity (or relative density) s of liquids. Fig. (25) shows a hydrometer in two liquids. It
has a stem of prismatic cross section of area (a).
In fluid 1 : W =
In fluid 2 : W = ( )
Hence, we can show that.
……………….. (2.46)
Where ( = Submerged volume in fluid 1 )
Note:
1. Equation (2.46) is used to divide the scale of the stem after specifying the geometry
and weight of the hydrometer.
2. Hydrometer is made usually from a glass or Perspex tubes. If the weight of these tubes
do not satisfy the hydrometer equation (W = ), then additional weight, such as mud,
lead, is added inside the bulb.
3. For each hydrometer, there are minimum and maximum values of (s) which can be
measured by its scale.
2.6 Stability of Submerged and Floating Bodies
2.6.1 Submerged Bodies
A completely submerged (immersed) object has a "rotational stability” when a
restoring couple is set up by any small angular displacement. This occurs only when the
center of gravity (G) is below the center of buoyancy (B), see Fig. (26)
Rotational Stability
Fig. (26)
Fig. (25)
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2.6.2 Floating Bodies
I Liner Stability; A floating body has a linear stability when a small liner displacement
in any direction set up restoring force tending to return it to its original position. A
small upward displacement decreases the volume of liquid displaced, resulting in an
unbalanced downward force, and small downward displacement results in greater
buoyant force, which causes an unbalanced upward force.
II Rotational Stability; Any floating object with G below B floats in stable equilibrium.
Certain floating objects, however, are in stable equilibrium even when G above B.
This may occurs when the line of buoyant force after the heel intersects the center line
above G, at a point called the "Metacenter M", see Fig. (27), which is a prismatic
body (all parallel cross – sections are identical). When the body is displaced by angle
( ) clockwise, a counter clockwise restoring couple is set up, whose value is;
Restoring Couple = W * MG sin ..................…. (2.27)
Fig. (27)
The distance (MG) is called the "Metacentric Height", and it is direct measure of the
stability of the body. The following criteria are used for this purpose;
MG > 0 {M above G} Stable Equilibrium
MG < 0 {M below G} Unstable Equilibrium ……… (2.48)
MG = 0 {M and G coincides} Neutral Equilibrium
For a general floating body of variable cross section, such as a ship, a convenient
formula can be developed for determination of the metacetic height (MG). This will be
done in the next article.
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2.6.3 Metacenter
The (Metacenter M) is the intersection of the buoyant force after the angular
displacement and the original center line; or, it is the intersection of the lines of action of
the buoyant forces before and after the heel (angular displacement).
In what follows, we will derive a formula for determining the Metacentric height (MG).
This will be done for a ship, which is a non – prismatic body (parallel cross – sections are
not identical), see Fig. (28). The results are, however, general and can be used for bodies
with prismatic cross – sections.
The Metacentric height (MG) is found as;
MG = MB
BG . ………….… (2.49)
() sign is used when G above B
(+) sign is used when G below B
Now; MB =
……………………………. (2.50)
Where; x= Centroidal distance of the displaced volume after the heel (B')
To specify (x), we take moments of volume about the line (00);
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( )
=∫
∫
Fig. (28).
𝛼 𝑥
o
o
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=∫
But d =x tan dA
Thus; x =∫
n A
And, hence;
x =
∫ A n
…………………………… (251)
Where: ∫ Second moment of area of water line surface about longitudinal
axis (oo)
= Displaced (submerged) volume
Thus, (2.51) in (2.50) gives;
MB =
……………………….. (2.52)
And hence (2.52) in (2.49) gives
MB =
……………………….. (2.53)
After calculating (MG) from equ. (2.53), the criteria specified in equ. (2.48) is used to
specify the state of equilibrium of the body.
2.7 Relative Equilibrium
In this application, the fluid is in motion such that no layer moves relative to an
adjacent layer; as a result, the shear stresses throughout the fluid are zero (absent), which
is the same as in fluid statics. Therefore, the laws of fluid static are applied for this type of
motion. Examples are;
1. A fluid moves with uniform velocity
2. A fluid moves with uniform linear acceleration
3. A fluid with uniform rotation
2.7.1 Uniform Linear Acceleration
A liquid in an open vessel is given a uniform linear acceleration (a=axi+ayj), see
Fig. (29). After some time the liquid adjusts to the acceleration so that it moves as a solid;
i.e, the distance between any two fluid particles remains fixed, and hence no shear stresses
occur. We will apply Newton's 2nd
law to find the pressure gradients and distribution
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Fig. (29)
∑ x
(p
) (
)
…(2.54)
∑ y=m
(p
) (
)
Thus;
( +g) …………………………. (2.55)
Equations (2.54) and (2.55) will reduce to equations (2.3) and (2.5) if (ax=0 and ay=0)
respectively. Now, to find the change in pressure;
dp =
dp =  ( ) …………………………… (2.56)
The angle of inclination ( ) for the lines of constant pressure can be found from
equ. (2.56) by setting (dp=0); thus;
0 = ( ) y
Hence;
………………….. (2.57)
Equation (2.56) can be integrated with B.C. (p= 0 at x=0, y=0) to obtain;
p = 0 ( ) ………………….. (2.58)
)p+𝜕𝑝
𝜕𝑦
𝑑𝑦
)𝑑𝑥
)p𝜕𝑝
𝜕𝑥
𝑑𝑥
)𝑑𝑦 )p+
𝜕𝑝
𝜕𝑥
𝑑𝑥
)𝑑𝑦
)p𝜕𝑝
𝜕𝑦
𝑑𝑦
)𝑑𝑥
dw=𝜌𝑔𝑑𝑥𝑑𝑦
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Equation (2.55) can be integrated in analogous way to equ. (2.7), to obtain the following
formula;
= + ( ) …………………………. (2.59)
Again, for (ay=0), equ. (2.59) reduces to equ. (2.7) For static fluids
Note; For open tanks ( 0 ) and point (o) is at the middle (center) of the width of
the tank. For closed tanks ( 0 ) and point (o) is not at the middle (center) of the
width.
1.7.2 Uniform Rotation
Rotation of a fluid, moving as a solid, about an axis is called" Forced Vortex
Motion". Every particle of the fluid has the same angular velocity. This motion is to be
distinguished from" FreeVortex Motion", in which each particle moves in a circular path
with speed varying inversely as the distance from the center. A liquid in a container, when
rotated about a vertical axis at constant angular velocity ( ), see Fig (30), moves like
solid after some time interval. No shear stresses exist in the liquid, and the only
acceleration that occurs is directed radially inward towards the axis of rotation (an=
and ) Thus, analogous to equations (2.54) and (2.55), we can wirte;
( ) ………….. (2.60)
( ) …………… (2.61)
The pressure distribution analogous to equ. (2.58) will be;
= 0+
( ) ……………..(2.62)
For points along the horizontal plane (y=0), equ.(2.62) gives;
0
0
……………..(2.63)
Which gives the distance to the free surface.
In a similar way, equ. (2.61) can be integrated to obtain;
= + ( ) ………………………. (2.64)
Where: h= Vertical downward distance from the free surface
Fig. (30)
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Note:
1. Lines of constant pressure are of parabolic type.
2. Volume of the parabola =
(Volume of cylinder with same base and height)
3.
0
0
Thus; z =
Examples
Example (2.1): For the system shown in the figure, calculate the manometer reading (H).
Sol.:
60000+0.2 13.6 0.92 =16000
H= 0000 0 0 0 000
∴ 0.3242 m= 32.42 cm
Example (2.2): Determine the pressure difference between the water pipe and the oil pipe
shown in the figure, in Pascals and in meters of water.
Sol.:
0
0 ( )
∴ 0
0 0
0 O
Example (2.3): If the weightless quarter cylindrical gate shown in the figure is in
equilibrium, what is the ratio between ( ) and ( )?
Sol.:
F= hA=
R=
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= =
∑
F*(R )= ( )
∴
Example (2.4): For the twodimensional weightless solid body shown in the figure,
determine the magnitude and direction of the moment (M) applied at the pivot required to
hold the body in the position shown in the figure.
Sol.:
= =9810*1*2=19620 N
y
1.332 m
=
30819 N
0.849 m
F= =9810*1*2.828=27746.9 N
=1.414+ ( )
=1.885 m
∑ =0
F*0.47+
∴M=13078 N.m
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Example (2.5): A (2cm) diameter cylinder of wood (s=0.5) floats in water with (5 cm)
above the water surface. Determine the depth of submergence of this cylinder when placed
in glycerin (s=1.25). Will it float in stable, unstable or neutral equilibrium in this case?
Sol.:
In water: W=
( 5)
Thus ; h=10 cm
In glycerin:
∴ x=4 cm
5cm , =2cm
00=
MG= 00
cm
< ∴Unstable Equilibrium
Example (2. 6): Calculate the weight and specific gravity of the object shown in the
figure to float at the wateroil interface as shown.
Sol.:
W=
=
( )
( )
=
=9810*0.00159
∴W=15.6 N
s= ⁄
=
( 0 )
0 =
000
∴s=0.899
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Example (2.7): The cylindrical vessel shown in the figure is rotated about its vertical
longitudinal axis. Calculate:
1. The angular velocity at which water will start to spill over the sides.
2. The angular velocity at which the water depth at the
center is zero, and the volume of water lost for this case.
Sol.:
(a)
∴ H = 2z = 150mm
H =
=0.15
(
)
∴ = 34.4 rad/sec
(b) H'= (
)
0.3=
(0 )
rad/sec
=1.178*10
3 m
3
∴
Example (2.8): An open cylindrical tank (0.9m) high and (0.6m) in diameter is two –
thirds filled with water when it is stationary. The tank is rotated about its vertical axis,
calculate:
1. The maximum angular velocity at which no water is to spill over the sides.
2. The angular velocity at which the bottom of the tank is free of
water for a radius of (150mm).
Sol.:
a H =
*0.9 = 0.6m
∴ m
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( )
∴ =11.43 rad/sec
b z
( )
( )
W = 16.16 rad/sec
Example (2.9): For the closed rectangular tank shown in the figure, calculate the pressure
at points (A) and (B), and the forces on sides (AB) and (CD).
Sol.:
tan =
0 º
tan
………………..(1)
…………………….. (2)
(1)& (2) gives:
y = 0.814 m and x =1.77 m
p2=p1+ (ay+g) h
=140000 + 0.68*1000(0 + 9.8) h
Thus; p2 =140000 + 6670.8h
z = 2.4 tan z = 0.288 m
pA = 140000 + 6670.8*z pA = 141.92 kPa
pB = 140000 + 6670.8*(1.2 + z) pA = 149.93 kPa
To calculate the hydrostatic forces, we have to imagine a free surface at which (p=0 gage)
above the existing surface at which (p=140 kPa):
Z' = 0000
0 0 z' = 21m
FAB= hA=0.68*9810*(0.6+z+z')*1.2*1.5 FAB=262.6 kN
FCD= hA=0.68*9810*(0.386+z'0.6)*1.2*1.5 FCD=249.6 kN
A
B
D
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Problems
The problems number listed in the table below refers to the problems in the
"textbook", chapter"2".
Article
No. Related Problems
2.2+2.3 4,5,7,10,13,14,16,20,21,22,24,26,29,30,31,32,33,34
2.4 36,37,39,40,44,45,46,53,59,60,61,62,63,65,68,69,71,78,79,82,8
6,89,91,101,106
2.5+2.6 92,93,96,97,98,100,102,103,105
2.7 110,112,114,115,116,117,118,122,123,124,126
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Fig. (1)
Chapter 3
Fluid Flow Concepts
3.1 Definitions and Concepts
Velocity (V): It is the time rate of change of displacement of fluid particles. It is a vector
quantity:
( )
.................... (3.1)
Acceleration ( ): It is the time rate of change of velocity vector.
( )
( )
Thus;
a=
And;
Where;
Streamline (S.L.): It is an imaginary line or
curve drawn in the fluid flow such that the
tangent drawn at any point of it indicates the
direction of velocity ( ) at that point. Since the
Local
Acceleration
Total
Acceleration
Convective Acceleration
…………..(3.2)
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Fig. (2)
Fig. (3)
velocity vector has a zero component normal to streamline, there can be no flow across a
streamline at any point, see Fig.(1). Streamlines indicate the direction of motion in
various sections of fluid flow.
Types and Classification of Flow
1 Internal and External Flow
Internal Flow; is bounded by a wall (surface) around all the
circumference of flow. Examples are pipe or duct flows, flows
between turbine or compressor or pump blades see Fig (2).
External Flow; is bounded by a wall (surface)
from one side and free at other sides. Examples are
flow over a flat plate, over airfoil, over a car, over
airplane fuselage, see Fig. (3).
2 Steady and Unsteady Flow
Steady Flow; none of the flow and fluid variables (velocity, acceleration, density.....)
vary with time.
Unsteady Flow; any one of the variables change with time
Steady Flow: ( )
Unsteady Flow: ( )
3 Uniform and Non – Uniform Flow
Uniform Flow; velocity vector ( )remains the same at all sections of the flow.
Non  Uniform Flow; velocity vector ( ) changes from section to sections of the flow.
Uniform flow;
Non  Uniform flow;
𝑢∞ 𝑢∞
….................... (3.3)
s=space .........................(3.4)
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Fig. (4)
Fig. (5)
Fig. (6)
Fig. (7)
4 Laminar and Turbulent Flow;
Laminar Flow; fluid particles move in smooth paths in layers or laminas with one layer
sliding over an adjacent layer. The paths of
individual particles do not cross or interact see
Fig.(4). Laminar flows occur at low velocities (low
Reynolds number Re) , where (Re =
) ; =
characteristic length.
Turbulent Flow; paths of various particles are irregular and random with no systematic
pattern of flow see Fig.(5). Turbulent flow occurs at high velocities (high Re). In
turbulent flow, the velocity (u) fluctuates with time. The time average velocity ( ) can be
found as, see Fig. (6);
∫
0
where;
u = Instantaneous velocity
= Fluctuation from the mean value
T= Time period
Note: For pipe flow; Re < 2000 Laminar Flow
Re > 4000 Turbulent Flow
5 One, Two  and Three  Dimensional Flow
One  Dimensional (1D) Flow; the variation of
fluid and flow parameters transverse to the main
flow are absent. These parameters remain constant
at any cross  section normal to the main flow see
Fig.(7). Mathematically;
( ) ................................ (3.6)
……………….(3.5)
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Fig. (8)
Fig. (11)
Fig. (9)
Fig. (10)
Two Dimensional (2 D) Flow; the variation in the
flow and fluid parameters takes place in x and y
directions only see Fig.(8). This means that all the
particles move in parallel planes (xy planes) along
identical paths in each plane. there is no variation
normal to this plan. Mathematically;
( ) ........................................(3.7)
Three  Dimensional (3D) Flow; the flow and fluid
parameters have variation in all the three directions,
See Fig.(9).Mathematically;
( ) ......................................(3.8)
6 Viscous (Real) and Non Viscous (Ideal) Flow
Viscous (Real) Flow; effects of viscosity exist and
cause reduction of velocity inside the boundary
layer (b.l.)
Non  Viscous (Inviscid, Ideal); effects of viscosity
are absent outside the (b.l.), see Fig.(10).
7 Incompressible and Compressible Flow
Incompressible Flow; the flow in which the density ( ) is assumed constant
( = constant ). Examples are flow of liquids and gases with low velocities (M ).
Compressible Flow; the flow in which the density ( ) is not constant, but varies with
pressure and temperature. Examples are gas flow and special types of liquid flow (such as
water hammer phenomena).
Volume Flow Rate (Q) and Mass Flow Rate ( )
Volume Flow Rate(Q); is the volume rate of fluid
passing a section in a certain fluid flow. Thus (see
Fig. (11));
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Fig. (12)
Q=
Or
Q= A.V = AV A V
..........................(3.9)
Where;
A=Cross  sectional area of the flow.
V= Average velocity of the flow.
A.V =AV cos = AV cos(0) =AV
Mass Flow Rate ( ) ; is the mass rate of fluid passing a section in a certain fluid flow;
Or; kg/s ……….....(3.10)
If the velocity (V) is variable across the section, then;
Q= ∫
…………..... (3.11)
= ∫
…………..... (3.12)
EX: For laminar flow through pipes, the
velocity distribution across the section is given
by;
(
) , see Fig. (12)
Thus;
Q = ∫ ∫ ( (
)
) ×
0
Hence; Q =
= AV, where V= average velocity
The average velocity (V) can be calculated as;
V =
∫
∫ ( (
)
) ×
= ∫
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Fig. (13)
System (sys)
It is a quantity of matter of fixed mass and identity. the system boundaries may be
fixed or movable. System exchanges energy only with the surrounding
Control Volume (C.V.)
It is a fixed region in the space bounded by the control surface (C.S.). The control
volume (C.V.) can exchange both mass and energy with the surrounding.
Note; System approach is usually used in solid mechanics, where the body is clearly
identified and can be followed during its motion. In fluid mechanics, a "system" of fluid
cannot be easily followed during its motion, since its boundaries are not clear. Instead, a
“control volume" approach is used, where a fixed volume specified in the fluid is
considered and the changes in this c.v. due to flow of fluid system through it is studied.
3.2 General Control Volume Equation
The general control volume equation, which will be derived in this article, relates
the changes in a general property of the "System" (Mass, Energy, Momentum) to that
occurred in a “Control Volume" through which the system flows, see Fig. (13).
Let;
N = Total amount of a general property within the system (mass, momentum, energy)
And;
= The amount of (N) per unit mass (N/m)
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Thus;
N = m = = ∫ ∫
And;
= ∫
∫
Now, the rate of change of (N) within the system is;
l
0
) )
l
0
(∫ ∫
)
(∫
)
By adding and subtracting the term(∫
) ;
l
0
(∫ ∫
)
(∫
)
l 0
(∫ ∫
)
= l 0 ) )
l 0
( )
=
=
∫ ∫
∫
=
∫ ∫
Hence, the final form of the “General Control Volume Equation“ is;
∫ ∫
...................... (3.13)
Equation (3.13) will be used to derive FOUR laws of conservation that are used to
analyze any fluid flow problem. These are:
1 Law of Conservation of Mass: Continuity Equation (C.E.)
2 Law of Conservation of Energy: Energy Equation (E.E.)
3 Law of Conservation of Liner Momentum: Momentum Equation (M.E.)
4 Law of Conservation of Angular Momentum: Angular Momentum Equation.
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Fig. (14)
3.3 Law of Conservation of Mass: Continuity Equation (C.E.)
The property (N) is the mass (m) of the system. Thus;
N= m
[ mass cannot be created nor can be destroyed]
Thus; from equ.(3.13);
∫ ∫
General C.E. ......….…….... (3.14)
For the flow control volume shown in Fig.(14);
∫
( ) ( )
( ) ( )
=
Hence, equ.(3.14) can be written as;
.......………... (3.15)
For steady flow, (
), thus;
∑
∑
C.E. for steady flow …………………..(3.16)
For incompressible flow ( ), hence;
∑
∑
C.E. for steady incompressible flow ………..(3.17)
3.3.1 Continuity Equation at a Point
To derive the continuity equation at a point, not for a C.V. (equ.3.15), we will
apply this equation for an infinitesimal cubic element of fluid, see Fig.(15).
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Using equ.(3.15) ;
+(
) (
)
Substituting for each term and simplify, it is obtained;
( )
( )
( ) General C.E. at a point ……….(3.18)
For incompressible flow ( )
=0,
Thus, equ.(3.18) becomes;
+
C.E. at a point for incompressible flow …………………(3.19)
3.4 Law of Conservation of Energy: Energy Equation (E.E.)
The property (N) is the energy (E) of the system, which has three forms; internal
energy (Eu), potential energy (E P.E) and the kinetic energy (E K.E). Thus;
N = E sys = Eu + EP.E. + Ek.E.
= mu + mgz +
And;
(𝜌𝑣 𝜕
𝜕𝑦 𝜌𝑣
𝑑𝑦
)𝑑𝑦𝑑𝑧
𝜌𝑤𝑑𝑦𝑑𝑥 𝑑𝑦
…………………………….(3.20)
Fig. (15)
(𝜌𝑢 𝜕
𝜕𝑥 𝜌𝑢
𝑑𝑥
)𝑑𝑦𝑑𝑧
𝑚
(𝜌𝑢 𝜕
𝜕𝑥 𝜌𝑢
𝑑𝑥
)𝑑𝑦𝑑𝑧
𝑚
(𝜌𝑤 𝜕
𝜕𝑧 𝜌𝑤
𝑑𝑧
)𝑑𝑦𝑑𝑥
𝑚
𝑚
𝜌𝑢𝑑𝑦𝑑𝑧
𝜌 𝑑𝑥𝑑𝑧
(𝜌𝑣 𝜕
𝜕𝑦 𝜌𝑣
𝑑𝑦
)𝑑𝑥𝑑𝑧
𝑚
(𝜌𝑤 𝜕
𝜕𝑧 𝜌𝑤
𝑑𝑧
)𝑑𝑦𝑑𝑥
𝑚
𝑑𝑥
𝑑𝑧
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……………………………………..(3.21)
Thus; equ.(3.13) becomes;
∫
∫
1st law of thermodynamics:
Q  W = dEsys ……………………………………… (3.22)
Where: (Q=Heat) and (W=Work). Hence;
∫
∫
……………(3.23)
Now, the work (W) has two components; shaft work (Ws) and flow
work ( ∫
); i.e.;
W = Ws + WF
= Ws+∫
= Ws+∫
+∫
Hence, equ.(3.23) becomes;
∫ ∫ (
+e) General E.E. …………(3.24)
For steady flow,(
∫ )
; thus;
∫ (
) …………………………….(3.25)
Using (3.21); and for one inlet an one outlet (
)
*(
) (
) + ………..(3.26)
Now;
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Heat per unit mass
( ) work per unit mass
Where hT = Turbine head (m)
hp = Pump head (m)
Substitute (3.27) in (3.26) gives;
q  (
) (
) …………………..(3.28)
Equ.(3.28) is the Steady State Steady Flow Energy Equation (S.S.S.F.E.E). In
differential from, it can be written as;
dq d = d
+ du + gdz + d
=
dq d =du+pd
…………………..(3.29)
From thermodynamics;
du+ pd =Tds …………………(3.30)
and;
Tdsdq=d(ghL) ……………..(3.31)
Where;
s=Entropy (J/kgK)
hL=Losses head (m)
Thus; equ.(3.29) becomes;
+ d(ghL)=0 ………….(3.32)
Integrate equ.(3.32) between sections (1) and (2) for incompressible fluids ( )
using equ.(3.27); it is obtained (after dividing by g);
(
) (
) ……….. (3.33)
………………………….(3.27)
Available Energy
at section (1) Energy
Added Available Energy
at section (2)
Energy
Subtracted
Energy
Lost
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Fig. (16)
Equation (3.33) in the S.S.S.F.E.E. for incompressible fluids. It is applied between
two sections (1) and (2) along, the fluid flow, see Fig.(16).
3.4.1 Bernoulli's Equation (B.E.)
It is a special from of the E.E (3.33) along a streamline without losses (hL=0)
and with no shaft work (hp=0,hT=0). Thus, B.E. will be;
B.E. …………………………(3.34)
B.E. is applied for steady frictionless flow of incompressible fluid along a stream line
Notes for Application of equations (3.33) and (3.34) (E.E &B.E.)
1 Available energy of the fluid has three forms;
Pressure (flow) energy per unit weight (m)
Kinetic energy per unit weight (m)
Potential energy per unit weight (m)
2 Each term of the E.E. and B.E. represents energy per unit weight (head h (m)).
3 The energy of the fluid can be represented by four forms;
a Total energy E = p
(J)
b Energy per unit weight h =
(m) (N. m/N)
c Energy per unit volume
(N/m
2)
d Energy per unit mass
(J/kg)
4 In equs. (3.33) and (3.34), section (1) is upstream and section (2) is downstream.
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5 (hp) in the E.E. represents the energy per unit weight added to the fluid by a pump,
compressor, fan, blower. (hT) represents the energy per unit weight subtracted from
the fluid by a turbine or windmill.
6 To calculate the power (P);
P =
Thus; P = gh =
7 (V) in the E.E. and B.E. is the average velocity at the section.
8 (hp) in the E.E represents the output power (OP) of the pump. To calculate the input
(shaft) power (IP);
……………………(3.36)
Pump efficieny
9 (hT) in the E.E. represents the Input Power (IP) to turbine. To calculate the output
power (OP);
=
……………………..(3.37)
Turbine efficiency
10 Kinetic Energy Correction Factor ( ):
The K.E. term (
) in the E.E. does not represent the average of the K.E.
across the section. Instead, we should calculate (
) across the section, where (u)
is the variable velocity across the section. To account for that, we should use (
) in
the E.E, where is the K.E correction factor, calculated from;
∫
∫
Hence;
=
∫( (
) ……..………….(3.38)
For laminar flow in pipe (
(
) ),
For turbulent flow in pipe
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3.5 Law of Conservation of Linear Momentum: Momentum Equation (M.E)
This law represents the application of the Newton's 2nd
law for a system. The
property (N) is the linear momentum of the System (mv); that is;
∑
( ) ............…………………… (3.39)
The R.H.S. of this equation is obtained from the general control volume equation (3.13).
Thus, with;
N = (mv) linear momentum
=
Equ.(3.13) becomes;
( )
∫
∫
..................………….(3.40)
Substitute (3.40) in(3.39), the general momentum equation (M.E) is obtained;
∑
∫
∫
∑ ( )
( ) ( )
Since the force (F) and the momentum ( ) are vector quantites, equ.(3.41) may be
written in three directions as;
∑ = ( )
( ) ( ) …………..(3.41a)
∑ = ( )
( ) ( ) …………..(3.41b)
∑ = ( )
( ) ( ) …………..(3.41c)
For steady flow,( ( )
), and the general M.E. reduces to ;
∑ ( ) ( ) M.E for steady flow ..............….. (3.42)
Also,
∑ =( ) ( ) …………..(3.42a)
∑ =( ) ( ) …………..(3.42b)
∑ =( ) ( ) …………..(3.42c)
For problems with one inlet and one outlet, ( ) Thus, equs.(3.41)
and (3.42) become;
............…….…...(3.41)
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∑ ( )
( ) …………..(3.43)
∑ ( ) …………….(3.44)
This can be written in three directions as;
∑ ( )
( ) ………….(3.43a)
∑ ( )
( ) ………….(3.43b)
∑ ( )
( ) ………….(3.43c)
And;
∑ = ( ) …………..(3.44a)
∑ = ( ) …………..(3.44b)
∑ = ( ) …………..(3.44b)
3.5.1 Applications of Momentum Equation
3.5.1.1 Fixed Vanes (Blades)
Fixed vanes (or blades) are usually used to deflect a jet of fluid to a certain
direction ( ), see Fig.(17).
The momentum equation is used to derive expressions for the two Forces (Fx) and
(Fy) required to hold the blade stationary. The following assumptions for the problem are
used;
1 Steady flow.
2 Frictionless flow.
3 Incompressible fluid.
4 Change in P.E. is neglected (Zin=Zout)
Fig. (17)
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5 Atmospheric pressure at inlet and outlet
Applying the B.E. between the inlet and outlet gives;
The momentum equations (3.44a) and (3.44b) will be applied;
∑ ( )
 ( ) ( ) n l n ne
∑ ( )
Fy= ( ) n l n ne
The resultant force (F) is calculated as;
F=√
with
3.5.1.2 Moving Vanes (Blades)
3.5.1.2.1 Single Moving Vane
The vane here is moving with a speed (u), and the fluid jet is approached with
velocity (V1). To make this problem similar to that of fixed vanes, we use the principle of
relative motion, that is, assuming the vane fixed and the fluid is approached with relative
velocity (V1u), see Fig. (18)
Fig. (18)
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The M.E is used to find ( ) and ( ) required to obtain the desired motion (with
velocity u). The M.E.(3.44) is applied on the relative model, thus;
∑ ( ) ( )……………….(3.45)
With;
Thus, we can show that;
( ) ( ) ( ) …………….. (4.46)
Fy= ( ) sin …………………. (3.47)
Power Delivered by the vane =
………… (3.48)
K.E. remaining in the jet =
……………… (3.49)
3.5.1.2.2 Series of Moving Vanes
A series of vanes are fixed on a wheel, and thus a rotational motion is obtained.
Compared to the single moving vane, where part of the mass flow rate ( ) can reach the
moving vane, for a series of moving vanes the entire mass flow rate ( ) will be used,
thus, the M.E. will be;
∑ ( ) ( ) ……………..(3.50)
Where;
=
And, thus;
Fx= (1cos ) ( )( ) ……………(3.51)
Fy= n ( ) n ……………(3.52)
Power delivered by the vanes= Fx=
........……..(3.53)
K.E. remaining in the jet=
......………..(3.54)
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Fig. (19)
Note:
1 The most important application for a series of moving vanes are the turbine wheels,
pumps, compressor, fans,
2 If the blade moves towards the jet, then:
Vr=V1+u
Power=
v2>v1
3 If (V1) and (u) are not inlive, (Vr) is found by adding them vectorially;
Vr=V1 u
3.5.1.3 Forces on Pipe Bends
The momentum equation is used to
calculate the forces (Fx) and (Fy) required
to hold the vertical pipe bend shown in
Fig.(19). Equs. (3.44a) and (3.44b) is
applied here;
∑ ( )
( ) ( )
∑ ( )
( ) ( )
Where Wb=Bend Weight
Wf=Fluid Weight= ( =bend volume)
Notes:
1 Before applying the M.E, the C.E. and E.E. are usually applied between sections(1)
and (2) to find the unknowns that are not given with the data, such as (V2,p2,…..).
2 If the plane of the bend is horizontal, then;
a (z1=z2) and (z) will not be used in the E.E.
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Fig.(20)
b (Fx & Fy) will be in the horizontal plane
c (Wb & Wf) will not appear in the yM.E
d There will be a M.E. in the zdirection (equ.3.44c), with;
∑ [no flow in the zdirection]
3 If there is more than inlet or outlet (such as in Ybend),
we must use equs.(3.42a,342b and 342c).
3.5.1.4 Theory of Propellers
Propellers are used to generate thrust against the drag exerted by the fluid on a body
moving through it. The thrust is generated by increasing the momentum of the fluid
passing through it, see Fig.(20).
Applying M.E., equ.(3.44a);
F= (  )= (  )= (  )=( ) …………(3.55)
Where A=
Area of propeller
Thus; (  ) ……………………(3.56)
Applying B.E. from 1 & from 3
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Fig. (21)
Since p1=p4, hence;
p3p2=
(
) ……………………..(3.57)
Equating equations (3.56) and (3.57), and simplification yields;
…………………….(3.58)
Output Power OP=FV1= ( ) …………………….(3.59)
Input Power IP=
(
)…………………(3.60)
K.E remaining in the slip stream=
( )
……………(3.61)
IP=OP+K.E. remaining in slip stream
The theoretical mechanical efficiency ( ) is;
( )
(
)
……………….(3.62)
If , then;
……………………………..(3.63)
( ) is maximum when ( ) is minimum, which occurs when ( )is large i.e.,
accelerating larger Q to a smaller ( ) for a given ( )
3.5.1.5 Jet Propulsion
I Jet Engines
The thrust of the jet engine (Fth) is generated
by expanding the gases produced from burning a
fuel with air in the combustion chamber, to a
higher velocity (v2<v1), see Fig.(21) Applying
M.E,equ.(3.42a) gives;
Fth =(
)
or; Fth= ( ) ………………….(3.64)
Where f =
fuelair ratio
Z1=Z2
Z3=Z4
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Fig. (22)
Fig. (23)
II Boats
The pump of the boat accelerates the jet
of water from ( ) to ( ), see Fig(22).
Applying M.E. gives;
F= ( ) …………………..(3.65)
3.5.1.6 Rocket Mechanics
The rocket motor carries its fuel and oxidizer with it, and so, there is no mass flow
enters the motor, only exit mass flow rate exists (gases produced by burring the fuel and
oxidizer at a rate of ), and hence producing the thrust of the rocket motor ( ) The
mass of the rocket is changing with time, therefore we
have to apply the general M.E (equ.3.41b), see Fig (23);
R = Air resistance
mR = Mass of rocket body
mf = Fuel mass
Rate of fuel buring
Vr= Exit gas velocity relative to rocket
V1= absolute velocity of the rocket
V= rocket velocity w.r.t. a reference moving with the rocket
V=0 but
Rocket acceleration
Total mass m = mR + mf = mR + ( )
=( mR + mf)
=mi  …………………….(3.66)
Where;
Initial mass of the fuel
mi= Initial mass of the rocket
M.E.;
∑ ( )
( ) ( )
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Euler's Equation
1 Friction flow ………………(3.71)
2 Along a s.l.
Fig.(24)
Rmg= ( )
( )

Thus;
( )
……………………………(3.67)
( )
…………….(3.68)
Where
Fth= thrust of the rocket ………………..(3.69)
3.5.2 Euler's Equation of Motion
This equation is applied for frictionless flow along a streamline. To derive this
equation, we will apply Newton's 2nd
law the fluid element shown in Fig.(24).
∑
Since V=V(s,t), then;
a=
=
……….(3.70)
Thus;
pdA(p+
) 
(
)
Simplification yields;
3.5.3 Bernoull's Equation
Euler's equation (3.71) for steady (
) incompressible fluid ( const.)
yields;
Integrating this equation;
𝜃
(p+∂
∂ ) A
𝑐𝑜𝑠𝜃 𝑑𝑧𝑑𝑠
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Fig.(25)
Or;
=
3.6 Some Applications of Continuity, Momentum and Energy Equations
3.6.1 Losses due to Sudden Expansion
The head loss due to sudden expansion (he) from a C.S.A. (A1) to a larger C.S.A
(A2) is found by applying the three continuity, momentum and energy equations. The loss
in energy occurs due to formation of a separated region after the sudden expansion, see
Fig.(25) This region is a low pressure
region accompanied by the formation of
vortices. The following assumptions are
made to analyze the problem;
1 Steady incompressible turbulent flow.
2 Uniform velocity over the cross –
sections.
3 Shear forces on the wall are negligible.
4 At the sudden expansion section (1'), the lateral acceleration of the fluid particles in
the eddy along the surface is small, and so, hydrostatic pressure variation occurs
across the section (p=p1).
C.E: A1V1=A2V2 …………………………(a)
E.E:
=
……………………… (b)
M.E: p1A2p2A2= ( )
………………...(c)
Solution of equs. (a), (b) and (c) simultaneously for (he) gives;
he =( )
(
) …………………(3.73)
For circular pipes;
he =* (
) +
…………………(3.73a)
Bernoulli's Equation
1 Steady flow
2 Incompressible fluid …………..(3.72)
3 Frictionless flow
4 Along the same s.l.
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Fig.(26)
3.6.2 Hydraulic Jump
A rapidly flowing stream of liquid in an open channel suddenly changes to a
slowly flowing stream with a larger C.S.A. and a sudden rise in elevation of liquid
surface. This phenomenon is called"
hydraulic jump". It converts the K.E to P.E.
and losses or irreversibilities (rough surface
of the jump), see Fig.(26). The following
assumptions are used;
1 Steady incompressible flow.
2 Uniform velocity across the C.S.A. A1&A2.
3 Shear forces at the wall are negligible.
4 Hydrostatic pressure distribution across the sections (1) &(2).
The continuity, momentum and energy equations are applied to the problem to find the
head loss (hj) due to hydraulic jump, and the new elevation (y2).
C.E.: A1V1=A2V2
A1y1.1 =A2y2 .1 ………………..(a)
E.E:
=
y1+
y2
……………….(b)
M.E: ∑ ( )
F1 – F2
………………….(c)
From (a) and (c);
√(
)
…………………(3.74)
From (b);
( )
……………(3.75)
3.7 Law of Conservation of Angular Momentum: Moment of Momentum Equation
The property (N) of the system here is the moment of linear momentum (mrxv).
This law represents the application of the Newton's 2nd
law of motion for a system;
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Fig.(27)
∑
( rx )
rxF=
( ) …………………..(3.76)
The R.H.S of this equation is obtained from the general control volume equation (3.13).
Thus; with;
rx
= rxv
Equ.(3.13) becomes;
( )
∫
∫ ( )( )
……..(3.77)
Hence, equ. (3.76) becomes;
r x F =
∫
∫ ( )( )
......……..(3.78)
Equ. (3.78) is the General Moment of Momentum Equation. The most important
application of this equation is for turbo machines, in which the normal (n) and tangential
(t) coordinates are used, see Fig.(27). The C.V. is the annular flow bounded by the root
circle (r1) and the tip circle (r2) of the blades.
The flow is assumed steady. The torque (T)
on the (C.V.) is produced from the tangential
components of forces (Ftr). Thus, with;
rxv=rVt
v.dA=VndA=Q=Q2=Q1
Equ.(3.78) reduces to;
Ftr=∫
=∫ ∫
Or;
T= ( ) ( ) Euler's Equation for Turbomachines ….(3.79)
For problems with more than one inlet and outlet, this equation becomes;
T=( ) ( ) .......................................…. (3.80)
Torque
on C.V.
Rate of Change of Moment
of Momentum in C.V. Net Efflux of Moment of
Momentum from C.V.
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Examples
Example (3.1): A twodimensional reducing bend has a linear velocity profile at section
(1). The flow is uniform at sections (2) and (3). The fluid is incompressible and the flow
is steady. Find the magnitude of the uniform velocity at section (3).
Sol.:
C.E.:
∫
0
0
*
+0
=1.35+0.45
∴ ⁄ ( ve sign means inflow at 3 )
Example (3.2): Water enters a two
dimensional channel of constant width
(h), with uniform velocity (U). The
channel makes ( ) bend that distorts
the flow to produce the velocity profile
shown at the exit. Evaluate the constant
(c) in terms of (U).
0
v = c ( 3.5 – x / h )
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Sol.:
C.E.: A1V1=A2V2
Uh1=∫
0 ∫ (
0)
Uh= *
+0
Uh=c*
+
Uh=3ch
∴ c=
m/s
Example (3.3): For the water tank shown in the
figure, calculate the exit velocity (V2) required to
lower the water level in the tank by (10 cm) in
(100 seconds).
Sol.:
Q1=A1V1=
( )
General C.E.:
( )
Thus;
( )
But;
0
00 m/s (level decreasing)
Thus; ( ) ( )
( )
=(0.00785+0.01) +0.001
( )
m3/s
0.7267=
( )
3
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Example (3.4): Neglecting the losses, calculate the required pump shat horsepower.
Sol.:
B.E. 12: 0+0+0=
+
…………….(1)
Man. Equ.: P2+0.6 +0.175*13.6 =0
0.6 – 0.175 * 13.6 = 2.98,
Sub. in (1) V2= 3.372 m/s
C.E.: A2V2=A3V3
V3= 23.98 m/s
E.E. 13: 0+0+0+hp= +( )
+0
∴hp=31.74 m
∴ kW =49.1 h.p.
Example (3.5): A horizontal axisymmetric jet of air ( kg/m3) with (10mm)
diameter strikes a stationary vertical disk of (200mm) diameter. The jet speed is (50 m/s)
at the nozzle exit. A manometer is connected to the center of the disk. Neglecting the
losses and the difference in potential head, calculate:
a The deflection (h) of the manometer.
b The force exerted by the jet on the disk.
c The thickness (t) of the air jet at the exit.
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Sol.:
B.E. 12
( )
Man.equ.:
∴ cm =0.0888m
∑ ( )
F=1.22*
( )
∴ n l
n
B.E. 13:
V1=V3
Qin= Qout
∴t=0.125 mm
Example (3.6): At what speed (u) should the vane shown in the figure travel for
maximum Power? What should be the angle ( ) for maximum power?
Sol.:
∑
( )
Fx= 0 ( cos  )
Thus;
Fx= 0 ( cos )
or
Fx= 0( 0 ) ( cos )
𝜌
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Power P=Fxu= 0( 0 ) ( cos )
For maximum Power
0( cos ) ( 0 ) ( 0 ) =0 u=
0
0[( 0 ) ( ) ]=0 =0 min. power
max.power
Example(3.7): An air plane traveling (400 km/h) through still air ( N/m3),
discharge (1000m3/s) through its two (2.25m) diameter propellers . Dertermine;
a The theoretical efficiency b the thrust c The pressure difference across the propeller,
and d The theoretical power required
Sol.
a V1=400
000
00 m/s
V=
000
( )
m/s
b F= ( )
( )
V=
Thus; F=
( )
c F=( )
000
( )
=4.53 kpa
d =
000
0 kW
Example (3.8): Determine the burring time for a rocket that initially has a gravity force
of (4.903 MN), of which (70 percent) is propellant. It consumes fuel at constant rate, and
its initial thrust is (10 percent) greater than its gravity force. The exhaust gases velocity is
(Vr=3300 m/s). Considering (g) as constant at (9.8 m/s2), and the flight to be vertical
without air resistance, find the speed of the rocket at burnout time, its height above the
ground, and the maximum height it will attain.
Sol.:
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Fth=1.1Wi=1.1*4.903 Fth=5.393 MN
Fth=
0
00 kg/s
Mass of propellant = 0 0
Hence;
The burring time=
= 0
buringtime=214.35 s
a=
( )
= 0 0
kg
Thus;
a=
00 00 0
00 0
Simplification gives;
Thus;
V ln( ) n
When t=0, V=0 ( )
Thus;
V ln(
0 )
at burnout, t=214.3, thus;
V ln(
0 )
The height at t= 214.3; (yb) is
yb=∫ ∫ * ln(
0 )+
0
0
y
The rocket will glide (V2/2g) higher after burnout, thus
ymax= yb+
=117.2*103+
y x
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Problems
The problems number listed in the table below refer to the problems in the
"textbook", chapter"3"
Article
No. Related problems
3.3 6,7,8,10,12,13
3.4
+
3.4.1
17,18,20,21,22,24,26,27,28,30,31,32,33,34,35,36,38,39,41,42,46
,47,51,52,53,54,55,58,59,60,61,63,64
3.5
+
3.5.1.3
70,76,77,78,82,84,85,87,88,89,125,137,138,139
3.5.1.1
+
3.5.1.2
100,111,112,113,114,116,118,120,121,122,123
3.5.1.4
+
3.5.1.5
91,92,93,95,97,99
3.5.1.6 101,102,105,106,107,108
3.6 127,128,130,133,134,136
3.7 140,141,142,143,144,145,146
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Chapter 4
Dimensional Analysis and Similitude
4.1 Introduction
The solution of most engineering problems involving fluid mechanics rely on data
acquired by experimental means. The fullscale structure employed in the actual
engineering design is called the "PROTOTYPE", while the replica of the structure on
which tests are made is called the "MODEL", which is usually made much smaller than
the prototype for economic reasons.
The process of manufacturing the model, and the conversion of the experimental
results obtained from the model tests, both are based on laws and relations obtained from
“Dimensional Analysis” and “Similitude”, sees Fig. (1).
Fig. (1)
4.1.1 Dimensions and Units
"Dimensions" are physical variables that specify the behavior and the nature of a
certain system, whereas the "Units" are used to specify the amount of these dimensions.
In fluid mechanics problems, the "fundamental" dimensions are three; these are
Mass (M), Length (T) and Time (T). The dimensions of any physical variable include
these three dimensions (one, two or all). In heat transfer and thermodynamic problems, a
fourth dimension for temperature ( ) is added. Table (1) includes the dimensions of the
most important variables.
Prototype
Actual Full
Scale Structure
Dimensional
Analysis &
Similitude
Model
Replica of the
Structure
Experimental
Results
Experiments
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Table (1)
Quantity Dimension Quantity Dimension
Mass M Dynamic Viscosity (μ) ML1
T1
Length L Kinematic Viscosity (ν) L2 T
1
Time T Density ML3
Force MLT2
Pressure ML1
T2
Area L2 Specific Weight ML
2 T
2
Volume L3 Surface Tension MT
2
Velocity LT1
Power ML2 T
3
Discharge L3T
1 Work ML
2 T
2
Angle M°L°T° Angular Velocity T1
Notes:
1 Any analytically derived equation should be dimensionally homogeneous, i.e., the
sum of exponents for each dimension in the R. H.S. of the equation should be equal
to that in the L. H. S. of the equation. Also, the dimensions of each term in any
equation should be the same.
Examples:
1) Q = AV
L3T
1 = L
2. LT
1
L: 3 = 2 + 1 = 3
T: 1 = 1
2) P = γh
ML1
T2
= ML2
T2
.L
M: 1 = 1
L: 1 = 2 + 1 = 1
T: 2 = 2
Empirical equations need not be dimensionally homogeneous, for example, the
Chezy formula;
√ c = constant
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√
The equality of dimensions is absorbed in the constant of the equation (c).
2 In fluid mechanics problems, the maximum number of dimensions in any physical
quantity, or group of physical quantities is three dimensions (m=3), i. e., (m) could be
(0, 1, 2 or 3 (max)).
3 Physical variables in fluid mechanics are usually classified in to three groups;
1) Geometry Variables (L, d, A, V, Angle,)
2) Fluid Properties Variables ( , μ, ν, k, σ,)
3) Kinematics and Dynamic Aspects of Flow Variables (p, v, a, g, F, ω, P, W,)
4 Any variable which has no dimension is called "Dimensionless" variable; with
dimensions (M°L°T°).
4.2 Dimensional Analysis
The basic objective in dimensional analysis is to reduce the number of separate
variables involved in a problem to a smaller number of independent dimensionless groups
of variables called “Dimensionless Parameters”. As the functional relationship between
the variables can be analytically established (by using dimensional analysis), the
experimental work is reduced to finding the value of the constants. Dimensional analysis
can be gainfully used to plan the experiment and reduce the number of variables to be
tested; and the results can be presented meaningfully.
Consider, for example, the motion of a sphere in a fluid. The drag force (FD)
excreted by the fluid on the sphere is known to depend on the sphere diameter (D),
velocity (V), fluid density ( ) and viscosity (μ), see Fig. (2). If we want to study the
effect of the independent variable (P, M, V, D) on the dependent variable (FD), we have
to investigate the effect of each of the four variables, separately and keep the other three
constants. Thus, if, for example, we made (5) experiments for each variable, we need
(54 = 625) experiment to study these effects and relations, which is a very complex and
laborious process, since it involves the use of different fluids, different diameters and
different velocities. Besides; the results presentation is a very complex and cannot be
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representative. To solve this problem, we use the principle of dimensional analysis, by
which we replace the original relation;
( )………………. (4.1)
to the following relation:
(
)
Or ( )
We should note the following;
1 Number of variables in the original problem (equ. 4.1) is (n = 5).
2 Number of “dimensionless” variables in the transformed problem (equ. 4.2) is (2),
which is the difference between the number of variables in the problem (n = 5) and the
number of dimensions of the problem (m = 3).
3 The number of experiments in the original problem is (54 = 625), whereas in the
transformed problem it is (51 =5).
4 We note that ( , V, D) are repeated with (CD) and (Re) in (equ. 4.2). These are called
the repeated variables of the problem, and their number equals to the number of
dimensions of the problem (m = 3).
5 During the (5) experiments, we have to change (Re) five times and measure the drag
force (FD) and then draw the relation between them. The change of (Re) can be done
by changing one of the four independent variables ( , V, D, μ) and keep the other
three constants.
6 The obtained results (CD = f (Re) relation) can be used for different ranges of (P, V, D,
M) that used in the experiments.
7 The process of finding the dimensionless parameters (CD & Re) from the original
variables (FD, , V, D, μ) is made by using the “Buckingham  Theorem”, which
will be described in the next article.
4.2.1The Buckingham – Theorem
“If a physical phenomenon involves (n) quantities, and if these quantities involve
(m) dimensions, then the quantities can be arranged into (nm) dimensionless parameter”.
..............................…………………….. (4.2)
CD = Drag Coefficient , Re = Reynolds Number
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Let (A1, A2… An) be the (n) quantities involved in the problem. A functional
relationship should exist such that;
( ) ……………………….. (4.3)
If the number of dimensions in the above (n) quantities is (m), then there exists (nm)
dimensionless parameters called ( ) such that;
( ) ........…………… (4.4)
Where each ( ) term is a nondimensional (dimensionless) product involving the above
quantities.
The method of determining the parameters is to select (m) of the (A) quantities
with different dimensions that containing among them the (m) dimensions and use them
as “Repeating Variables” together with one of the other (A) quantities for each ( ). Thus;
The exponents are to be determined such that ( ) is dimensionless.
The Procedure for the Application of the Theorem:
1 Select the (n) variables involved in the problem.
2 Write the functional relationship F (A1, A2… An) =0.
3 Identify the dependent variable of the problem.
4 Select the repeating variables;
a) They must contain the dimensions of the problem, collectively.
b) Each repeating variable must have different dimensions from the others.
c) They must be physically the most significant variables.
Ex.
d) They must not contain the dependent variable.
e) They must contain least dimensions (Ex. (v, g) choose v).
f) They should not form a  term by themselves.
………………….. (4.5)
(A1, A2, A3) are the repeating
variables
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g) It is essential that no one of the repeating variables be derivable from the others.
h) A variable of minor importance should not be selected as repeating variables.
5 Write the parameters in terms of the unknown exponents;
6 For each expression, writ the equations of the exponents so that the sum of the
exponents of each dimension will be zero.
7 Solve the equations simultaneously.
8 Substitute back into the  expressions of step (5 ).
9 Establish the functional relationship [f ( ) =0] or solve for one of the
( s ) explicitly; for example;
( )
10 Recombine, if desired to alter the forms of the  parameters, keeping the same
number of s [( ) (
) , c = constant].
4.2.2 Common Dimensionless Numbers
In a hypothetical fluid flow problem, the pressure difference ( ) between two
points in the flow field depends on the geometry of the flow
(l, l1, l2), fluid properties ( , μ, σ, K) and the kinematics of the flow (V and g). Thus;
F ( p, l, l1, l2, , μ, σ, K, V, g) = 0
n = 10
m = 3
Dependent Variable = p
Repeating Variables = , V, l
Or:
, which is called the ( Pressure Coefficient Cp )
No. of 𝜋 s = 10 – 3 = 7
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i.e.:
………………… (4.6)
× ×
×
(
)
Euler number (E) is usually defined as;
√ √
……………….. (4.7)
Or:
, which is called the ( Reynolds Number Re )
i.e.:
.....………………… (4.8)
Viscous Force
Thus; Re=
At high (Re), the effect of viscosity is small and the (Re) effect is negligible because the
intertie effect is large. High (Re) means either (v) is large or ( ) is small or (l) is large.
Or;
√ lle e ( )
i.e.;
√
y … (4.9)
Gravity Force Fg = weight =
Thus;
√
√
√
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( ) is used when there is density stratification, such as between salt water and fresh
water.
Or;
lle e ( )
i.e.;
… (4.10)
r e en n r e l
l
l l
Weber number (W) is used when surface tension effects are predominant, such as in
capillary tubes and capillary waves in channels.
Or;
√ lle ( )
i.e.;
√
..................… (4.11)
Where: C= speed of sound in fluid = √
Fc =
√
Mach number effects are important when (M > 0.3 or 0.4). (M) is a measure of the ratio
of K.E. of the flow to internal energy of the fluid.
… (4.12)
… (4.13)
are geometry parameters.
Hence;
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f (
)
Or;
(
) .....................................… (4.14)
Equation (4.14) includes all effects which could be involved in a certain fluid mechanics
problem. Usually, not all these effects exist with the same degree of importance, but one
of them (or two) usually overcome the other effects. The inertia effect, of course, exists
always in all applications, since it is responsible of the fluid flow. The pressure effect is
the dependent parameter of the problem.
4.3 Similitude
It is the theory and art of predicting the prototype performance from model tests
and observations. It involves the application of dimensionless numbers, such as (Re, Fr,
W, and M) to predicate prototype performance from model tests. The similitude involves
three types;
I Geometrical Similitude
The shape of the model is similar to the prototype. The ratio between all
corresponding lengths (including roughness elements) between model (m) and prototype)
(p) are equal. The ratio ( ) is called the "Scale Ratio" or "Model Ratio". ( is read as
"lambda")
Length Ratio .........................… (4.15)
Area Ratio .........................…. (4.16)
Volume Ratio ........................… (4.17)
II Kinematic Similitude
a The paths of homologous moving particles are geometrically similar.
Pressure
Effect
Viscous
Effect Gravity
Effect
Surface
Tension
Effect
Compressibility
Effect
Geometry Effect
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b The ratios of velocities and accelerations at homologous particles are equal.
Velocity Ratio
...............................… (4.18)
Acceleration Ratio
.......................… (4.19)
Discharge Ratio
..........................… (4.20)
III Dynamic Similitude
This implies that in geometrically and kinematically similar systems at homologous
points the ratio of corresponding forces between the systems are same. This implies that
the dimensionless numbers derived in article (4.2.2) are the same between model (m) and
prototype (p);
(
)
(
)
Hence from equ. (4.14);
(
)
(
)
Or; (
) (
)
Hence;
.................................. (4.21)
Equ. (4.21) is a general formula. To find the force ratio between model and
prototype we need to calculate the velocity ratio ( ), which is found by equating (Re, Fr,
W, M) between (m) and (p). But since these numbers have different forms, each one give
different value for ( ), so, we cannot solve the problem by equating all these numbers.
Instead, we choose one of them which have greater effect than the others on the problem.
Most engineering applications for incompressible fluids, the (Re) and (Fr) are the most
important numbers. Each number has certain applications, as will be explained soon.
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4.3.1 Re  Criterion
This criterion is applied for engineering applications in which only inertia and
viscous effects are important. These include the following;
1 Flow in closed conduits (internal flow).
2 Flow around immersed bodies.
3 Flow in wind and water tunnels.
The problem here is solved by equating (Re); thus;
(
)
(
)
Or;
........................ (4.22)
Substitute (4.22) into (4.21) gives;
Or;
............................ (4.23)
It is clear from equ. (4.23) that when the same fluid is used for the model and prototype
( ) , the force ratio is ( ), i.e. ( ). this result is for problems
using Recriterion only.
4.3.2 Fr  Criterion
This criterion is applied for engineering applications in which only inertia and
gravity effects are important. These include the following;
1 Open channel flow.
2 Dams and spillways.
3 Hydraulic jump.
4 Ships and boats (neglecting viscous effects).
Now;
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Frm = Frp
(
√ )
(
√ )
√
Or;
√ √ ............................................ (4.24)
Substitute (4.24) into (4.21) gives;
............................. (4.25)
4.4 Ships Models Tests
The ships are usually exposed to two components of drag. These are viscous
(frictional) drag (FDv), which is a (Re criterion), and the ware drag (FDw), which is a (Fr
criterion). Thus, the total drag (FDT) is:
And;
To find the velocity ratio ( ), we use equs. (4.22) and (4.24). Thus;
Re – Criterion:
Vm < Vp ( since < 1 )
Fr – Criterion: √ √ Vm > Vp
So, we cannot solve the problem by equating both (Re + Fr), since each number gives
different value. To solve the problem, we test the model according to (Fr criterion) and
use equ. (4.24) for ( ), and calculate the viscous drag (FDv) from existing relations. The
ship model is tested in the laboratory and (FDTm) is measured, with the velocity calculated
………………………………….. (4.26)
ReCriterion FrCriterion
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from equ. (4.24). The following procedure is used to calculate the total drag on the
prototype and the power required to drive it;
1 Make a model test according to Fr criterion, equ. (4.24), and measure (FDTm) in the
laboratory.
2 Calculate (FDVm) from analytical or empirical relation.
3 Calculate FDWm = FDTm – FDVm.
4 Using Fr criterion, equ. (4.25), calculate (FDWp), i.e., FDWp = FDWm
5 Calculate (FDVp) from analytical or empirical relations.
6 Calculate ( )
7 Calculate the required ( Shaft Power = / ηp ), where (ηp) is the propeller
efficiency.
Note:
1. All problems of ships models tests which consider the viscous (FDV) and gravity
(FDW) effects are solved by following the procedure described above (steps 1 – 7).
The simple difference is in the relation used to calculate (FDV), steps 2 and 5.
2. If we neglect the viscous effects (FDV) and consider the gravity effects (FDW) only,
then the problem is solved according to Fr – Criterion ( Article 4.3.2 ).
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Examples
Example (4.1): The drag force (FD) on a sphere moving in a fluid depends on the sphere
diameter (D) and velocity (V), and the fluid density (P) and viscosity (M). Find the
functional relationship for the drag force (FD).
Sol.
F (FD, V, D, , μ) = 0
n = 5
m = 3
Dependent variable = FD
Repeating variable = m = 3 = ( , V, D )
: x1
Vy1
Dz1
FD
: x2
Vy2
Dz2μ
: Mº Lº Tº = (ML3
)x1
(LT1
)y1
(L)z1
MLT2
M: 0 = x1 + 1
L: 0 = 3 x1 +y1 +z1 + 1
T: 0 = y1– 2
Thus;
= 1
V2
D2
FD =
= CD Drag Coefficient
: Mº Lº Tº = (ML3
)x2
(LT1
)y2
(L)z2
ML1
T1
M: 0 = x2 + 1
L: 0 = 3 x2 +y2 +z2 1
T: 0 = y1 – 1
Thus;
= 1
V1
D1μ =
MLT2
LT
1
L
ML3
ML1
T1
No. of 𝜋 s = nm = 53 = 2
Properties Geometry
Kinematics & Dynamic Aspects
x1 = 1
y1 = 2
z1 = 2
x2 = 1
y2 = 1
z2 = 1
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Or;
Reynolds Number
Thus;
f ( ) = 0 f (CD, Re) = 0
CD = f (Re) The function (f) is found from experiments
Example (4.2): A(1/5) scale model automobile is tested in a wind tunnel with the same
air properties as the prototype. The air velocity in the tunnel is (350 km/h), and the
measured model drag is (350 N). Determine the drag of the prototype automobile and the
power required to overcome this drag.
Sol.:
Use Recriterion (equs. 4.22 and 44.23).
Rem = Rep
Equ. (4.22):
Since the same air is used ( ), thus;
Equ. (4.23):
Power = 000
00 er
Example (4.3): The wave resistance of a model of a ship at (1:25) scale is (7 N) at a
model speed of (1.5 m/s). What are the corresponding velocity and wave resistance of the
prototype ? Assume the model is tested in fresh water ( = 1000 kg/m3) and the ship will
operate in the ocean where (s = 1.03) for the water.
Sol.:
Use Frcriterion (equs. 4.24 and 44.25).
Frm = Frp
Equ. (4.24): √ √ Vp = 1.5√ √ Vp = 7.5 m/s
Equ. (4.25): Fp = 7 * 1.03 * 1 * 25
3 Fp = 112656 N
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Example (4.4): A (10 m) model of an ocean tanker (500 m) long is dragged through
fresh water at (3 m/s) with a total measured resistance of (103 N). The surface drag
coefficient ( cf ) for model and prototype are (0.245) and (0.0147) respectively, in the
equation ( F = cf A V2 ). The wetted surface area of the model is (20 m
2). Find the total
drag on the tanker and the power required at the propeller shaft assuming an efficiency of
(90 %) for the propeller.
Sol.:
FDVm = cfm Am Vm2 = 0.245 * 20 * 3
2 FDVm = 44.1 N
FDWm = FDTm – FDVm = 103 – 44.1 FDWm = 58.9 N
Using Frcriterion;
Vp = Vm√ √ = 3 √ √ Vp = 21.2 m/s
FDWp = FDWp FDWp = 58.9 * 1 * 1 * (500/10)
3 FDWp = 7362500 N
( 00
0) Ap = 20 * 50
2 Ap = 50000 m
2
FDVp = cfp Ap Vp2 = 0.0147 * 50000 * 21.2
2 FDVp = 330338.4 N
= 7692838.4 N
er er
Problems
The problems number listed in the table below refer to the problems in the
"textbook", Chapter "4”;
Article No. Related Problems
4.1 5,6
4.2 2,8,10,11,12,13,17,20
4.3 18,21,23,26,33,35
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Fig. (5.1)
Chapter 5
Viscous Fluid Flow
5.1 Introduction
In viscous flow, the viscous (frictional) effects are very important and must be
considered. Real (viscous) fluid flow may be either laminar or turbulent. In laminar flow
(low Re) the influence of viscosity is predominant. In turbulent flow (high Re) the inertial
effects are predominant.
5.2 Equations of Motion for Viscous Flow
These equations must include
the following forces:
1. Pressure forces.
2. Inertia forces.
3. Viscous (shear) forces.
4. Body forces.
5. External forces.
For each plane:
(1) Normal stress ( ).
(2) Shear stresses ( ).
Thus;
6 surfaces 12 Shear forces
6 Normal forces
Also, there is; 3 Body forces ( ) per unit mass
3 External forces ( )) per unit mass
Hence; the total forces = 24 forces
Now; u=u (x, y, z, t), v = v (x, y, z, t) w = w (x, y, z, t)
( )
( )
( )
𝜏𝑥𝑦 𝜕𝜏𝑥𝑦
𝜕𝑥 𝑑𝑥
𝜍𝑥 𝜕𝜍𝑥𝜕𝑥
𝑑𝑥
𝜏𝑥𝑧 𝜕𝜏𝑥𝑧𝜕𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑦
𝑑𝑧
𝜏𝑥𝑦
𝜏𝑥𝑧
𝜏𝑎𝑏
𝜍 a : axis normal to the plane
b : direction of stress
𝜍𝑥
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Stoke's law;
(
)
( )
(
)
( )
(
)
( )
(
) ( )
(
) ( )
(
) ( )
For incompressible fluids, the continuity equation reduces to;
( )
Now, the equations of motion for viscous flow are the Newton’s 2nd
law. Thus;
∑ x ∑ y ∑
z ( )
Applying (5.11) on the cubical element (Fig.(5.1)), and using the relations (5.1) to (5.10),
simplification and rearranging yields the following equations of motion for viscous
incompressible fluids. They are known as “ Navier Stokes” equations;
(
) ( )
(
) ( )
(
) ( )
Local
Acclamation Convective Acceleration
Inertia forces
Pressure
Forces
Body
Forces
External
Forces Viscous forces
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5.3 Laminar Flow Between Parallel Plates: “Couette Flow”
Assumptions:
1. Steady incompressible flow (
)
2. One – dimensional flow (v=0, w=0) and (
)
3. Fully – developed flow (
( ) )
4. Newtonian fluid (
)
5. Parallel flow (
)
Applying the Newton’s law for the element of Fig.(5.2);
∑ ( ( ))
Thus;
(
) (
)
Using ( n =  dh/dx) and simplification yields;
( ) ……….....(5.15)
but;
(Newton's law of viscosity). Thus;
( ) …………...(5.16)
Integrate equ. (5.16) twice to obtain;
Fig. (5.2)
(𝑝 𝜕𝑝
𝜕𝑥 𝑑𝑥 )𝑑𝑦
(𝜏 𝜕𝜏
𝜕𝑦 𝑑𝑦 )𝑑𝑥
𝜏𝑑𝑥
n 𝜃 𝑑
𝑑𝑥
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( )
………...(5.17)
B.Cs. ; at y=0 u=0 →B=0
at y=a u=U →A=
( )
Thus; equ. (5.17) gives;
( )
( ) ………...(5.18)
The volume flow rate (Q) across the section is ;
∫
0
( ) ………...(5.19)
The shear stress
( )
( ) ………..(5.20)
The term
( ) in the above equation is constant for each flow. It can be calculated
by one of the following two methods;
1.
( )
( ) ( )
( ) ( ) > ( )
2.
( )
n ( )
Note; There are many other cases for equ.(5.17) with different boundary conditions;
these are :
1. Fixed plates (U=0)
( )
( )
2. Lower plate moving.
3. Upper plate moves upward when the shaded areas are equal, then (Q=0)
y = a u = 0
y = 0 u = 0
𝜏
y = a u = 0
y = 0 u = U
y = a u = U
y = 0 u = 0
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4. Both plates move in opposite directions.
5. Free surface flow
5.4 Losses in Laminar Flow
For steady laminar flow, the reduction in ( ) represents the work done on
fluid per unit volume. The work done is converted into irreversibilities (losses) by the
action of viscous shear. Thus, the net work done and the loss of potential energy
represents the losses per unit time due to irreversibilities.
Fig. (5.3): Work Done and Loss of P.E. for a Fluid Particle
Net Power = n ( )
Substitute for each term in equ. (5.21), simplifying and neglecting higher order terms and
dividing by the element volume, and using equ.(5.15), it is obtained;
(
)
………...... (5.22)
For flow between two fixed parallel plates;
(𝜏 𝜕𝜏
𝜕𝑦 𝑑𝑦
𝑧) (𝑢
𝜕𝑢
𝜕𝑦 𝑑𝑦
)𝑑𝑥
𝑢 (𝑝 𝜕𝑝
𝜕𝑥 𝑑𝑥
)𝑑𝑦
𝑢 (𝑝 𝜕𝑝
𝜕𝑥 𝑑𝑥
)𝑑𝑦
(𝜏 𝜕𝜏
𝜕𝑦 𝑑𝑦
) (𝑢
𝜕𝑢
𝜕𝑦 𝑑𝑦
)𝑑𝑥
𝐹 𝑉
𝐹 𝑉
𝐹 𝑉
𝐹 𝑉 𝑝 𝜏 𝑢 (x,y)
𝑑𝑦 𝑑𝑥𝑑𝑦
x
n𝜃 𝑑
𝑑𝑥
y = a u = U
y = 0 u = v
y = a 𝑑𝑢
𝑑𝑦 𝜏 = 0
y = 0 𝑢
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( ) n
( )
Thus;
∫ (
)
[
( )]
0
( )
Hence;
( ) …….... (5.23)
The power done by the fluid between (1) and (2) is;
∫
( )
( )
( )
Thus;
( ) . ……….(5.24)
For horizontal flow, ( )
….....(5.25)
Note:
Equations (5.24) and (5.25) can be derived using the energy equation from (1)to (2);
( )
Thus;
( )
( ) …………...…….... (5.24)
For horizontal flows;
( )
Thus;
………... (5.25)
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Fig. (5.4)
y=ar
dy = dr
5.5 Laminar Flow through Circular Tubes and Annuli
The same assumptions used in article (5.3) for flow between parallel plates are
used and applied here. The difference is that the yaxis is replaced by the r axis.
∑ ( ( r ( ))
(
) (
( ) )
n
Thus;
( )
( ) ….………….. (5.26)
Integrate once;
( ) ………………... (5.27)
But;
Hence, (5.27) becomes;
( )
Integrate;
( )
( ) ……………... (5.28)
Equation (5.28) is applied for both circular tubes and annuli the difference is in the
boundary conditions.
𝜏 𝜋𝑟𝑑𝑥 𝜕
𝜕𝑟(𝜏 𝜋𝑟𝑑𝑥)𝑑𝑟
(𝑝 𝜕𝑝
𝜕𝑥𝑑𝑥) 𝜋𝑟𝑑𝑟 𝜏 𝜋𝑟𝑑𝑥
𝑝 𝜋𝑟𝑑𝑟
𝑑𝑤 𝜋𝑟𝑑𝑟𝑑𝑥
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5.5.1. Circular Tubes: Hagen  Poiseuille Equation
The boundary conditions for equation (5.28) are;
( )
Thus:
( ) ………………... (5.29)
The maximum velocity ( ) is at (r=0);
( ) ………………. (5.30)
The average velocity (V) is;
∫
∫
( )
0
Hence;
( )
………... (5.31)
Thus; equ. (5.29) can be written as ;
(
) (
) ………... (5.32)
Now, the volume flow rate (Q) is;
∫
( )
0
... (5.33)
For horizontal tubes; h= constant;
( )
Hence
Hagen  Poiseuilli equation .... (5.34) (horizontal tubes)
Thus;
……………... (5.35)
And ; for horizontal tubes
………………... (5.36)
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The shear stress ( ) is calculated as ;
( ) ………...(5.37)
Notes:
1. The term
( ) is calculated as in article (5.3) for flow between parallel plates.
2. For flow through circular pipes and tubes;
Re < 2000 laminar flow.
2000 < Re< 4000 transition flow.
Re> 4000 turbulent flow.
3. Equations (5.29) and (5.32) are for fully  developed flow, for which (
), which
occurred far from the entrance of the pipe.
5.5.2 Annuli
The boundary conditions for equ.(5.28) are;
At r = a u = 0
r = b u = 0
Thus, we can show that;
( ) *
ln
+ ……….... (5.38)
∫
( ) *
( )
+ ……….. (5.39)
𝜏𝑤
𝜏
𝜏
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5.6 Boundary –Layer Flow
Boundary –layer (b.l.) is a very thin layer of fluid adjacent to the surface, in which
the viscosity effects are important. The flow inside the boundary –layer is viscous (real)
flow, while outside the boundarylayer (main stream) the flow is inviscid (frictionless,
ideal).
u
S= b.l. thickness
Turbulent b.l.Transition
zone
Laminar
b.l.
Viscous flowSt
Frictionless (ideal, in viscid) flowU
Sy
U
x
Fig.(5.5): Growth of b.l. along a Smooth Flat Plate
To derive the equations of boundary –layer, we will apply the M.E to a control volume of
the b.l over a surface.
∑ =
( ) ( ) ( )
For steady flow of incompressible fluid along one side of a
smooth plate, the only force acting is due to drag or shear at
the plate, since the pressure is constant across the b.l. and
around the periphery of the c.v. of Fig.(5.6). Thus, for unit depth of the plate;
∑ ∫ ( ) ( ) ( ) …..(5.40)
C.S. ( )
AD
BC ∫
0
∫
0
AB ∫
0  ∫
0
Hence, equ.(540) gives;
ℓ
Fig.(5.6)
Fx=∫ 𝜏𝑤𝑥
0𝑑𝑥
y
AU
B U
XCSkin – frictional dtag
x
Uu
y
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Fx=∫
0 ∫ ( )
0∫
0(
) …………. (5.41)
The velocity profile inside the b.l.(
)is usually given as;
(
) (
)=F( )
Where
&dy=
y=0
y=
Thus, equ.(5.41) becomes;
Fx=∫
0 ∫
0(
)
∫
0
∫
0
(
)
Hence;
∫
0(
) ………………. (5.42)
Equation (5.42) is the M.E for 2D steady flow along a flat plate (b.l. equation). The
problem of the b.l. is to find;
1. Boundary –layer thickness along the plate ( )
2. The skinfrictional force ( ∫
0 )
Equation (542) is a general equation applicable for both laminar and turbulent boundary
layers. The only difference is in the form of the velocity profile (
)
5.6.1 Laminar Boundary – Layer
The velocity profile inside the laminar b.l. is given by the following formula;
( )
0 =1 y> ………. (5.43)
(5.43) in (5.42) gives;
∫ (
)(
0
) =0.139
…….. (5.44)
At the wall;
0
0
(
)
Thus;
η
η
η
η
U
y U
uy
u
δ
U
u / U
1
ξ
ξ δ
y
1
u / U
1
ξ
ξ
y
δ
(5.43)
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…………. (5.45)
Equation (5.44) and (5.45) gives;
10.78
dx
Integrate with B.C (x=0, );
√
Substitute (5.46) into (5.45) gives;
√
……………………………….. (5.47)
The skinfrictional drag force ( ) is;
∫ √
0……………….. (5.48)
The skinfrictional drag coefficient ( ) is;
√
Note:
For Re<5* Laminar b.l.
5* < e < Transition
Re> Turbulent b.l.
5.6.2 Turbulent Boundary –Layer
The velocity profile is given by the 1/7th
power law;
(
) ⁄ = ⁄ ………………… (5.50)
(5.50) into (5.42) gives;
………………….. (5.51)
Equ.(5.50) cannot be used to find (
0) since it has no derivative at (y=0).
Therefore, an empirical formula for ( ) is used;
(
) ⁄ ………….. (5.52)
ere 𝑅𝑒𝑥 𝜌𝑢𝑥
𝜇
𝑢𝑥
𝑣………………………………… (5.46)
Where:
𝑅𝑒𝐿 𝜌𝑢𝐿
𝜇
𝑢𝐿
𝑣………………………………… (5.49)
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Equate (5.51) and (5.52), and integrate with B.C.(x=0, ), i.e., we assume turbulent
b.l. from (x=0); we get;
0
⁄ …………………….. (5.53)
From (5.52); (
) ⁄ ……………… (5.54)
And ∫
0 0
⁄
0…………. (5.55)
&
0 0
⁄ ……………………….. (5.56)
Note: If we take the laminar part of the b.l. into consideration; then:
0 0
⁄
00
…………………… (5.57)
BoundaryLayer Displacement Thickness ( )
It is the distance that the main stream (inviscid flow) is displaced due to the
existence of the boundarylayer;
∫ ( )
0 dy
Thus;
∫ (
)
0 dy
= ∫ (
)
0 ……………. (5.58)
5.6.3 BoundaryLayer Separation
The fluid inside the b.l. is affected by three forces;
1. The forward pull due to the momentum of the outer free moving main stream
(with velocity U).
2. The retarding viscous forces on the solid boundary (frictional effects).
3. The pressure forces due to pressure gradient along the boundary, which may be
either negative (or favourable
< ) or positive (adverse
> ) or zero
pressure gradient (
)
δ
L
t
x
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𝐹𝐷𝑝 𝑊𝑎𝑘𝑒
𝐹𝐷𝑓
𝐹𝐷𝑓 < 𝐹𝐷𝑝
Separation of the boundary –layer occurs when the resultant of the above three forces
become against the flow direction, and thus pushes the flow far away from the surface
and causes separation of the boundarylayer, see Fig.(5.7).
Notes:
1. Separation occurs when (
> ) only. When (
flat surfaces) or when
(
< ), no separation can occur.
2. The onset of separation is the point at which (
0 )
3. The region between the separation s.ℓ. and the boundary is called "wake" of the
body.
4. The "wake" of the body caused additional
"pressure or form drag" in addition to the
"skinfrictional drag"
FD = FD = Total Drag
5. In the design, we have to delay the separation point in order to reduce the wake size.
6. The separation point depends on;
a The body shape.
𝜕𝑝
𝜕𝑥<
𝜕𝑝
𝜕𝑥>
𝜕𝑢
𝜕𝑦 𝑦 0
𝜕𝑢
𝜕𝑦 𝑦 0 >
𝜕𝑢
𝜕𝑦 𝑦 0 <
U
p pmin actual
Ideal
Fviscous Fpress Ffree stream
Separation s.l.
wakeu=0
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b The roughness of the surface.
c The turbulence level in the free steam.
d The Reynolds number.
e Flow conditions.
5.7 Drag and Lift
The drag force (FD) is the force component parallel to the
relative approach velocity excerted on the body by the moving fluid.
The lift force (FL) is the fluid force component on a body at right
angle to the relative approach velocity
FD= …………………………………. (5.59)
<
>
>
The drag coefficient (CD) is defined as;
CD=
…………………………. (5.60)
Where;
A=projected area perpendicular to the flow ( U), except for
airfoils, where A=c*s
Generally;
(Re,Shape)
= (M,Shape)for compressible flow
See Fig(5.21) and Table (5.1) in your textbook
The lift coefficient (CL) is defined as;
CL=
………………… (5.62)
See Fig.(5.23) & Table (5.1) in your textbook.
Total (profile)
Drag Skin (frictional) Drag
(viscous effects)
Pressure (form) drag (wake effect)
……….. (5.61)
U
FL
FD
U
FL
FDα
Disc
Stokes
law
sphere
CD
C= chordS= span
S
C
Re
CL
CD
22Angle of attack
α
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𝐷 𝑟
𝑈
𝐹𝐷
𝐹𝐵
𝑊
𝐅𝐢𝐠 (𝟓 𝟖)
Stoke's Law
The flow of viscous incompressible fluid around a sphere with very small Re
(Re=
< ) has been studied by Stoke. He found that;
FD=6 ………….. (5.63)
To find the terminal velocity (U) of the sphere (at which a=0),
see Fig.(5.8);
∑
FD + FB = W
6
=
Thus;
U =
( ) …………………… (5.64)
To calculate the drag coefficient (CD) for this case;
CD =
Thus;
………………….. (5.65)
5.8 Resistance to Flow in Open and Closed Conduits
P
A
Z2ϴ
Z1
W
X
ɤAL
A=C.S.A.
P= wetted perimeter
sin ϴ= Z1 Z2
L
Fig.(5.9): Uniform flow in Conduits
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Consider the steady uniform incompressible flow in conduit of constant C.S.A. (A)
and wetted perimeter (p), see. Fig.(5.9). We wish to calculate the head loss due to friction
hf between sections (1) and (2).
E.E. 12:
………………….. (5.66)
C.E.: A1V1=A2V2
Since A1=A2=A ………………. (5.67)
M.E.: ∑ [since =0 (steady and v=const.)]
Thus;
( )
…………….. (5.68)
The wall shear stress ( ) is usually defined as;
*
…………….. (5.69)
Where Fanning friction factor (found experimentally, or theoretically)
Using (5.66), (5.68) and (5.69), we can show that;
………………. (5.70)
Where
hydraulic radius
For circular pipe,
Equation (5.70) is a general relation applicable for open and closed conduits, laminar and
turbulent flow, and for any shape of uniform crosssection.
The head loss per unit weight per unit length ( ) is defined as;
S=
…………………. (5.71)
Thus;
V=√
√ √ …………. (5.72)
Where;
C = √ ⁄ ; Equation (5.72) is called Chezy formula.
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5.8.1 Open Channels Flow
For incompressible, steady flow at constant depth in prismatic (constant C.S.A)
open channels, "Manning" formula is widely used, in which;
C=
⁄
……………………….. (5.73)
From (5.72);
V=
⁄ ⁄ ……………………………… (5.74)
Where;
Cm=(1) in SI units and (1.49) in U.S units
S= slope of the bottom of the channel=slope of the water surface
n= Absolute roughness coefficient, depends on roughness of the surface (see Table (5.2)
in your text)
The discharge Q=Av=
⁄ ⁄ …………………….. (5.75)
Note: For open channel; (p1=p2), thus;
( )
S=
( )
slope of the channel
5.8.2 Steady Incompressible Flow through Pipes
For pipes,
…………………………. (5.76)
Where;
f=Darcy friction factor
Knowing that (
) equ.(5.70) becomes;
…………….. (5.77) DarcyWeisbach equation
All quantities in equ.(5.77) except (f) can be measured experimentally. The Friction
factor (f) is known to depend on the following parameters;
f=f(V,D, ) ………………….(5.78)
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where;
measure of the size of the roughness projection (m)
measure of the arrangement or spacing of the roughing
m= from factor, depends on the shape of the individual roughness elements,
dimensionless
Since (f) is a dimensionless factor, it can be shown (using theorem);
f=f (
,
,
,m)
=f (Re,
,
,m) ………………..(5.79)
The term (
) is called the "relative roughness". The experiments of Nikaradse show that
for one value of (
), the (f Re) curve is smoothly connected regardless of the actual
pipe diameters (D). The experiments proved that for one type of roughness, the following
relation is applied;
f=f(
)………..(5.80)
Equation (5.80) is plotted as "Moody Diagram", see Fig (5.10) and (5.32) in your
textbook
103
104
105
106
107
108
Re
0.00001
0.005
0.02
2 4
Laminar
zone zone
crit ical Transit ing Turbulent zone
Complete Turbulent zone
f ≠ f (Re)
f ≠ f (Re)
Steel
Costion
Concted
wood
Type of pipe__
__
__
__
E/D
Smooth pipe
Blains ration
0.1
f
0.02
0.08
0.07
0.06
0.05
Fig.(5.10): Moody Diagram
f = f(Re,
f = 0
𝑅𝑒 ⁄
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For laminar flow, equation (5.35) is;
………. (5.35)
But *
+
Thus; with (5.77);
……………….. (5.81)
Thus, for laminar flow, (f) is independent of ( )
For turbulent flow in smooth pipe, Blasius relation is usually used;
f=0
⁄ ……….. (5.82)
Equation (5.82) is valid up to (Re=100000).
5.8.3 Simple Pipe Problems
They are the problems in which the pipe friction (hf) is the only loss exist. In these
problems, the variables are usually 6 (Q, L, D, hf, v, ). Three of these six variables are
generally given or may be determined, these are (L, v, ). Thus, three variables remained.
Accordingly three types of problems exist, which may be solved by using;
1 DarcyWeisbach equation 2.C.E. 3. Moody Diagram
Type "I": (L,v, Q, D), to find : hf
Procedure;
1. Calculate V=
2. Calculate Re=
3. Calculate
4. By using (Re,
), calculated (f) from Moody Diagram
5. Calculate hf =f
Type "II" Given:(L,v, , D), to find : Q
Procedure:
1. Assume f=f1
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2. Calculate (V1) from hf =
3. Calculate
4. Calculate
5. Using ( ,
) calculate new value (f2) from Moody Diagram.
6. Compare (f1) with (f2) and continue until convergence is attaint
7. Calculate Q=
Type"III" Given (L, ,Q) to find ( )
Procedure:
1. Assume f=f1
2. Calculate (D1) from =
3. Calculate
=
4. Calculate(
)
5. Using (
) calculate (f2) from Moody Diagram
6. Compare (f1) with (f2) until convergence
7. D=D1
Note: In place of Moody Diagram, the following explicit form of (f) may be used;
f =
* (
) (
0 +
Also, Colebrook formula may be used;
√ [(
) (
√ )]…………………(5.84)
Equation (5.84) is for commercial pipes in transtion zone
5.8.4 Secondary (Minor) Losses
They are the losses occurred in pipelines due to bends, elbows, joins, valves,…etc.
They are called secondary losses, while the frictional losses (hf) are called "primary
losses". Thus;
<
𝐷
5000 𝑅𝑒
………..(5.83)
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Total losses=Primary (Frictional) Losses +Secondary (Minor) Losses
= hf + hminor
=f
…………………. (5.85)
Where (K) is the loss factor for the minor losses. It is found experimentally, except losses
due to sudden expansion or contraction.
1. Loss due to sudden Expansion (he)
he=( )
* (
) +
……..(5.86)
Thus;
* (
) +
…………… (5.87)
If the expansion is form pipe to reservoir (
) , then
( ) and the complete kinetic energy (V2/2g) is converted
into thermal energy.
2. Losses due to sudden Contraction (hc)
The head loss due to sudden contraction (hc) may be
found using equ.(5.86). The proces of converting
pressure head to velocity head ((1) ( )) is very
efficient and hence the head loss from (1) to (0) is small
compared to the loss between (0) and (2). Thus;
hc= 0 0 ( 0 )
…………………(5.88)
C.E. 0 0=
or; 0 0
……… (5.89)
Where; Contraction Coefficient (found experimentally)
Thus;
=(
)
……………… (5.90)
And hence; (
)
0
①
②
V1V2
K= 1
▼
①
②
0
V2P2
V1
D1
Vena contract
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3. Head loss at Pipe Entrance
K= 0.5▼
Square intel
K= 0.5
▼
Round intel
K=0.01 – 0.05
▼
Reentrant intel
K=0.8 – 1
Note: see table (5.3) in your textbook for (k) values.
Equivalent Length (Le)
Minor losses may be expressed in terms of equivalents length (Le) of pipes that have the
same head loss for the same discharge thus;
k
………………….(5.88)
minor losses may be neglected when they are only (5%) or less than (hf). In general,
minor losses may be neglected when there is (1000D) length between each minor loss.
Note;
L L Le
V DD X V Q Q
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Examples
Example (5.1): The upper plate shown in the figure is moving to the right with (Vu=80
cm/s) and the lower plate is free to move laterally under the action of the viscous forces
applied to it. After steady –state conditions have been established, what velocity (Vℓ)
will the lower plate have? Take:(t1=2mm, t2=1mm, 0.1 Pa.s, 0.04 Pa.s).
Assume constant pressure for the two plates.
Sol.:
Since p= constant & , then:
u =
&
[
]
*
+
= 0.8*0
0
0 0
0
0
0
+
∴ cm/s
Example (5.2): In the pistoncylinder apparatus for pressure gage tester shown in the
figure, the piston is loaded to develop pressure of known magnitude. Calculate the mass
(M) required to produce (1.5 MPa(gage) in the cylinder, when
the piston moves with (0.02mm/s). Calculate also the leakage
flow rate of oil for these conditions. Assume steady uniform
flow and neglect the weight of the piston.
Sol.:
∑
p*
ℓ
er l e
Upper plate
Vu
V
t1
t2
Oil μ1
S1=0.9
Oli μ2 S2=0.9
τ1
τ2
S2=0.92
D
a
M
t1
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p*
………(1)
( )
[
]
=0 0 0 0
0 00 0
0 00 0
* 0
0 +
sub. In (1)
1.5*106*
( )
∴
Q = (
*
+)
= (0 0 0 0 00 0
(0 00 0 )
0 )
∴ ⁄ ⁄ ⁄
Example (5.3): Determine the direction of flow through the tube shown in the figure, in
which ( = 8000N/m3) and ( Pa.s). Find the quantity flowing in liters per second
and calculate the Reynolds number for the flow.
Sol.:
Let the elevation datum be taken at section2;
p1+ h1=200000+8000*5=240 kPa
p2+ h2=300000+8000*0=300000 kPa
Since (p2+ h2)>( p1+ h1)
∴ Flow from (2) to (1)
( )
( ) ( )
( )
( )
( )
V=
=0 0000
(0 0 )
⁄
Re=
0 0 0 000
0 0
Since (Re<2000), the flow is laminar.
ve
flow upward Q=0.00036
𝑚
𝑠
=0.0368L/s
1 p1 = 200 KPa
2 p2 = 300 KPa
10 mm ɸ
5m
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Example (5.4): For the hypothetical boundary –layer on the flat plate shown in the
figure, calculate:
a. The skinfrictional drag force on the top side per meter of width.
b. The drag coefficient of the top side.
c. The wall shear stress on the plate at the trailing edge.
d. The displacement thickness ( )of the boundary –layer at the trailing edge.
e. The displacement thickness ( )of the boundary –layer at the trailing edge.
Given constants: kg/m3 , =1.8*10
5 Pa.s
3 mm4
3
1
2
U=40m/s
B.L.edge
L=30 cm
Sol.:
a At section (3):
M.E.: Fx=( ) ( ) ( )
C.S. ( )
12
34 ∫
0 ∫
0
23  ∫
0  ∫
0
 x ∫
+ ∫
Fx= ∫
∫
=
∫
0
∫
0
=0.5
U
u
yδ
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= 0.1667 ( )
∴ 0.96 N on plate
b CD
( )
∴ 0.00333
c
0
0 0
0 00
∴ 0.24 Pa
d ∫ (
)
0dy=∫ (
)
0
=*
+0
=
∴ 1.5 mm
Example (5.5): A smooth flat plate (3m) wide and (30m) long is towed through still
water at (20 c) with a speed of (6m/s). Determine the drag on one side of the plate and
the drag on the first (3m) of the plate.
Sol.:
For water at (20 c), from Table (c.1): ( = 998.2 kg/m3) and (v = 1.007 * 10
6
). For
the whole plate;
× 0
00 × 0 Re = 1.787 *10
8 ; turbulent
0 0
0 = 0.0016
Drag force FD = CD *
PU
2 * L * b
= 0.0016 *
* 998.2 * 6
2 * 30 * 3 FD = 2587.3 N
Let (Retr = 5 * 105) at which transition to turbulent b.l. occurs;
Thus:
00 0 = 0.084 m
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For the first (3m);
00 0 Re = 1.787 * 10
7: turbulent
CD = 0 0
0 CD = 0.00255
FD = CD *
PU
2L * b
= 0.00255 *
* 998.2 * 6
2 3 * 3 FD = 412.7 N
Note: Ltr << L, so it can be neglected, and me assume the b.l. turbulent right from the
leading edge (x = O).
Example (5.6): How many (30m) diameter parachutes (CD = 1.2) should be used to drop
a bulldozer weighing (45kN) at a terminal speed of (10m/s) through still air at (100kPa)
and (20 C).
Sol.:
00000
( 0 ) = 1.19 kg/m
3
∑
W = FD + FB
n = 0.4
n = 1 parachute
Example (5.7):A jet aircraft discharges solid particles of matter (d = 10 mm) and (s =
2.5) at the base of stratosphere at (11000m). Assume the viscosity ( ) of air to vary with
(y) from see level as ( = 1.78*105
3.06*1010
y). Estimate the time for these particles to
reach the see level. Neglect air currents and wind effects.
Sol.:
Using Stokes law;
( ); and assuming <<
Thus;
Thus;
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∫ ∫
0
000
0
( 0 ) 0 ∫ ( 0 )
0
000
Thus; t = 1301189 seconds = 15.06 day
Example (5.8): Determine the discharge for a trapezoidal channel (see the figure),with a
bottom width (b = 3m) and side slopes (1 on 1). The depth is (2m), and the slope of the
bottom is (0.0009). the channel has a finished concrete linins.
Sol:
From table (5.2) p.p. (230); text book, n = 0.012
×
A = 10 m
2
√
0
= 1.1547
Thus;
0 0 ( ) ( ) Q = 27.52 m
3/s
Example (5.9): Determine the head loss for flow of (140 L/s) of oil (v= 0.00001 m2/s)
through (400m) of (200mm) diameter cast iron pipe.
Sol:
0
0 0 0000 Re = 89127
From Moody diagram, Fig (5.32) p.p. 237 in your textbook, ( = 0.25mm)
Thus;
0
00
With (Re = 89127) and (
= 0.00125), from Moody diagram f = 0.023
= 0.023 * 00
(0 ) (0 )
= 46.58
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Example (5.01): Estimate the cost per month required to treat and circulate the water of
the swimming pool shown in the figure. The circulation rate is (0.1 m3/s) through
(100mm) diameter smooth pipe. The pump efficiency is (80%) and (1kwh = 65 fils).
Include all losses and also take (V = 106
m2/s).
Sol.:
0
(0 )
∴
0
0
∴
0
0
( 000)0
∴
Apply the E.E. between two points on the free surface of the pool:
0 + 0 + 0 + hp = 0 + 0 + 0 + 0 + hℓ
hp = hℓ ∑
∑
= [0.009407* 0
0 + 0.5 + 5 +10+10+20+10+20*0.9+10]*
( )
∴
IP =
=
0 0
0
∴
No. of kwh = IP * t = 874.29 * 30 * 24
∴ No. of kwh = 629488.8 kWh
Cost = 629488.8 * 0.065
∴ Cost = 40916.8 I.D.
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Example (5.00): The pump of the piping system shown in the figure is used to supply
(1m3/s) of water to the uphill station. Take (f = 0.014) and include all minor losses.
Calculate:
a. The required pump shaft power knowing that its efficiency is (80%).
b. The operational cost (in I.D.) of pumping (10000m3) of water to the upper reservoir,
knowing that: (1kwh = 300fils).
Sol.:
a
∴ , = 7.96 m/s , = 14.15 m/s
∴ , ,
E. E.:
+
(
)
= ( )
( )
( )
+
( )
(
0 )
( )
∴
0 0
0
∴
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b
0000
∴ t = 10000 s = 2.78 hr
No. of kwh = IP * t = 2600 * 2.78
∴ No. of kwh = 7228 kwh
7228 * 0.3
∴ Cost = 2168.4 I.D
Problems:
The problems number listed in the table below refer to the problems in the "text
book" , chapter "5".
Article No. Related Problems
5.3 1,2,3,6,7,9,10,11,12,13,14,16,17
5.5 20,21,22,24,27,28,29,31,33
5.6 43,44,45,46,47,48
5.7 42,49,51,52,54,55,56,57,62,63,64
5.8.1 65,66,68,69,70,72,75,77,78,81,82
5.8.2 +
5.8.3 +
5.8.4
25,83,84,87,88,89,90,95,96,98,99,
100,101,103,105,106,108,109,110,112,114,
115,116,117,118,119,120,121,123,124,125,129
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Chapter 6
Fluid and Flow Measurements
The chapter includes the measurements method used for the most important fluid
and flow parameters, such as density, pressure, velocity, discharge, viscosity. All the
described methods involve the application of the basic fundamental principles covered in
the previous chapters.
6.1 Density Measurements
The most important methods are;
1. Weighing method; weighing a known volume ( )of the liquid;
=
⁄
………………. (6.1)
(w and ) measured
2. Buoyancy Principle: A solid object of known volume ( ) is weighed in air ( )
and in the liquid ( ) whose density ( ) is to be determined;
………………….. (6.2)
( ) measured
3. Hydrostatic Principle: Placing two immiscible liquids in a Utub, one of known
density ( ), the other of unkown ( ).
……………… (6.3)
( ) measured
6.2 Pressure Measurements
At first, we must signify between three types of pressure. These are total
(stagnation) pressure (po), static pressure (p) and the dynamic pressure ( ). The relation
between these three types is;
po= p + pd = p +
……………….(6.4)
The static pressure (p) is the pressure of the fluid "sensed "or" feeled" by an observer that
is moving with the fluid with the same velocity (v) in magnitude and direction, i.e, the
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observer must not disturb the flow. The total (stagnation) pressure (po) is the pressure of
the fluid when it is stopped and comes at rest (v=0). If you swim in a river moving with a
velocity (V), and you leave your body to move "freely" without resisting the flow, then
the pressure you "sensed" or "feeled" is the static pressure of the flow. If you stopped
yourself in the river, then the flow becomes "stagnant" at your body (v=0), and hence the
pressure you "sensed" or "feeled" is the total (stagnation) pressure. The difference
between the two pressures is the dynamic pressure ( =
), which is a result of the
kinetic energy of the fluid.
Pressure measurements include the measurements of "static" and "total" pressure
only. The dynamic pressure is calculated as the difference between the two.
6.2.1 Static Pressure Measurements
The static pressure may be measured either by a "piezometer" opening or by a "static
tube"
V
Piezometer openings
V
Static holes
Static Tube
Piezometer Opening; a number of openings (holes) are made on the periphery of the
tube, pipe or duct. These holes are connected through a "ring" around them with one
opening connected to the manometer. When the inner surface of the tube is rough, this
method is not effective, since the flow will be disturbed. In this case, the static tube is
usually used.
V
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Static Tube; it is a blunt tube with closed nose and a number of
openings around the periphery at a section far away from the nose
(L>14d); to ensure that the flow becomes paralell after the nose
curvature.
Note: The discrepancy ( ) between the true and the measured static pressure normally
varies as the square of the velocity of flow, i.e. ;
………………….. (6.5)
Where (C) is a correction factor determined by calibration
6.2.2 Total (Stagnation) Pressure Measurements
This pressure is measured by using the "Pitot tube",
which is an opennose tube. The flow at the openend
comes at rest. This point is called "Stagnation Point", at
which (p=p0) and (V=0).
6.3 Velocity Measurements
The velocity is measured is measured through the measurement of the dynamic pressure
( =
) . As was mentioned earlier, this pressure is calculated as the difference
between the measured total (stagnation) and the static pressures. Thus;
= 0
0 √
0
…………………….. (6.6)
The "Pitot tube" and the " piezometer opening" or the static tube" are connected to the
same manometer to measure the velocity directly, as follows;
1. Combined Pitot Tube and Piezometer Openings
B.E. 12 =√
……………. (a)
Manometer equ.:
(
0
) ……………… (b)
Stagnation point (Po v=0)
L
d
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Thus;
√ ( 0
)………………… (6.7)
2. Pitot Static Tube ; which combines the
"Pitot" and "static" tubes. The same relation
for the velocity (6.7) is applied here. A
correction factor (C) must be introduced into
equation (6.7) to take into account the losses
and errors. Thus, the actual velocity (Va) is
Va = C√ ( 0
) ………………… (6.8)
(C) is found by calibration
Note:
The " Pitot Tube" alone may be used to measure the velocity, as follows;
B.E. 12:
thus
√
but p2= ( )
p1=
Hence;
V=√
or ;
√ …………………. (6.9)
A correction factor (c) may also be used with this equation to calculate the actual
velocity;
Va = C Vt=C√
v
s
So
R
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6.4. Discharge Measurements
6.4.1 Orifice Meters
There are two types for this meter. One is used for reservoirs and the other for pipes.
6.4.1.1 Orifice in Reservoirs
B.E. 12:
Thus;
√ ………………….. (6.10)
The actual velocity ( ) is found by defining
the velocity coefficient ( ) as;
=
………………… (6.11)
Hence;
√ ……………. (6.12)
The actual discharge ( ) is given by;
………. (6.13)
Where;
A2= jet area at the venacontract. (A2) is related to the orifice area (A0) through the
coefficient of contraction (Cc) by;
Cc =
0 ……………….. (6.14)
Thus; √ 0 0√
or;
0√ ……………….. (6.15)
Where;
……………… (6.16) discharge coefficient
…………….. (6.17)
Where; 0√ and
………………….(6.18)
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(Cd) may be found from equation (6.15) by measuring 0 H and ⁄ . Thus,
determination of either ( ) or ( ) permits the determination of the other from equation
(6.16). Several methods exists;
1. Trajectory Method; by measuring (x0,y0), the actual velocity ( ) can be determind
if the air resistance is neglected.
0 0
 0
(
0
)
0
√ 0 ⁄ …… (6.19)
But √
Hence
2. Direct measurement of ( ) with a Pitot or Pitot –static tube placed at the vena
contract (point 2)
3. Direct measurement of jet diameter by outside calipers
4. Momentum Method
∑ ( )
( )
But 0 × 0
Thus;
0
0………………….(6.20)
Losses in Orifice ; To calculate the losses in orifice, apply the E.E. between (1) and (2),
which gives;
ℓ
(
)
(
)
Hence;
ℓ ( )
(
)……………….(6.21)
0
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Borda Mouthpiece
∑ x ( )
0
but; 0√
hence;
0 0√ √
Thus;
1=2 …………….. (6.22)
Unsteady Orific Flow from Reservoirs Time of Emptying Reservoirs
The determination of the time to lower the reservoir surface a given distance is an
unsteady flow case of some practical interest. If the reservoir Surface drops slowly
enough, the error from using B.E. is negligible.
C.E.
Thus;
or;
but
0√
hence;
0√
∫
√ 0
∫
⁄
Where (AR) is a known function of (y). If (AR= constant), then
0√ (√ √ ) ………………. (6.23)
0
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6.4.1.2 Orifice in Pipes
It is used to measure the discharge in pipes at law Re (low velocities). It is simply a
plate with central hole (orifice). To calculate the discharge (Qa);
B.E. 12:
………… (1)
C.E. A1V1=A2V2=Cc A0 V2
0 ……………. (2)
(1) and (2) gives;
√
(
0
) ………………….. (3)
Manometer Equation gives;
(
0
) ………….. (4)
Thus;
√ (
0 )
(
0
) ………………… (5)
And;
√ (
0 )
(
0
) ………………… (6)
The actual discharge (Qa) is;
Qa = 0 0 0
Hence;
Qa = 0√ (
0 )
(
0
) …………………… (6.24)
Note;
The orifice in a pipe causes vortices to be formed behind the orifice plate. These
vortices represent losses. The losses increase as the velocity increases. Therefore the
orifice in a pipe is used for low Re. At higher Re, Venturi meter is used.
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6.4.2 Venturi Meter
It is used to measure the discharge in pipes at high
velocity (high Re). The contraction coefficient is unity,
thus;
Cc=1
………………… (6.25)
B.E. 12:
…………….. (1)
C.E.
…………… (2)
(1) and (2) gives;
√ *
+
(
) ……………….. (3)
Manometer equ.:
(
0
) …………… (4)
Thus;
√ (
0 )
(
) …………………….(5)
Thus;
√ (
0 )
* (
) + …………………….. (6.26)
Thus; ( ) depends on ( ) regardless of the orientation of the venturi meter, whether it
is horizontal, vertical or inclined.
In the venture meter, no vortices is formed and the losses in energy is reduced, therefore
it can be used for high velocities.
S0
R
K
h
1
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6.4.3 Rotameter
This device is used to measure the flow rate based on the princi ple that the drag force on
a body depends on the velocity over it. A "notched bob" is placed in a conical container
through which theflow passes over the bob. Applying force balance on the bob;
FD+FB=W
…………………… (1)
Where;
=Drag Coefficient
=Mean velocity in the annular space
Fluid density
Body density
Bob volume
Thus; Bob frontal area =
=√
(
) ………………… (6.27)
and;
Q=A √
(
) …………………. (6.28)
Where;
A=
( ) ……………………. (6.29)
A=Annular area
a= constant indicating the tube taper=2y tan
6.5 Viscosity Measurements
6.5.1 Rotating Concentric Cylinders
This method is based on Newton's law of viscosity;
Flow
w
Vm Vm
dFD FB
ƿ
y
D
ϴ
y
L
motor w
a
b
μ
dissStrings pointer
Torque
measurement
r1
r2
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When the outer cylinder is rotated, the motion is transmitted to the inner cylinder by the
action of viscosity ( ) of the fluid, thus excerting a torque (T) on it. This torque (T) is
measured by measuring the deflection angle ( ) rotated by the inner cylinder,
T=K …………………… (6.30)
Where K= stringe constant (N.m/deg)
This torque (T) equals the sum of the torques excerted on the cylinder walls (Tc) and
bottom (Td). Hence
T=Tc+Td
K =
∫ ∫
0
0
Or
K = *
+ ……………………… (6.31)
Thus, to measure the viscosity ( ), we need only to measure the deflection angle ( ),all
other parameters are known (specification of the device)
6.5.2 Capillary Flow Method
This method is based on the application of the HagenPoiseullif equation (5.34) for
steady, laminar incompressible flow.
Q=
…………………….. (6.32)
Q= ⁄
(Q,H) measured
( ) known
Thus, from equation (6.32), viscosity ( ) is measured.
6.5.3 Saybolt Viscometer
This method is also based on capillary flow and Hagen
Poiseuille equation. It is used to measure the kinematic
viscosity ( ). The time (t) necessary to drain ( ml)
is recorded. (t) in seconds is the saybolt reading.
h
=60 m
Capillary tube
Diameter = D
Length = L
Const bath
temp.
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Q=
But Q=
and
Thus;
………………..(6.33)
Where h=head in the reservoirA correction term is added to equ.(6.33) totake into
consideration that the flow is not fully developed. Thus;
………..(6.34) (C1,C2)= constants of the device
𝑣𝜌
𝚤
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Examples
Example(6.1): A (75mm) diameter orifice under a haed of (4.88 m) discharges (907.6
kg) of water in (32.6s). The trajectory was determined by measuring (X0=4.76m) for a
drop of (1.22 m).Determine Cv,Cc,Cd, the head loss per unit gravity force, and the power
loss.
Sol.:
√ =√ m/s
t=√ 0 √
t=0.499s
0
0 m/s
⁄
0 000⁄
m
3/s
0 0√
( )
The head loss=H(1 ) =4.88(1 ) loss=0.241
Power loss= Q* Loss =9810*0.0278*0.241 Power loss=65.7 W
Example (6.2): A tank has a horizontal C.S.A. of (2m2) at the elevation of the orifice,
and the area varies linearly with elevation so
that it is (1m2) at a horizontal crosssection
(3m) above the orifice. For a (100mm)
diameter orifice, (cd=0.65), compute the
time, in seconds, to lower the surface from
2.5 to 1 m above the orifce.
Sol.:
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A=a+by
Thus;
A=2
y………………………(1)
C.E:
0√ =0
Hence;
t=
0√ ∫ ⁄
Using equ.(1);
t=
0
(0 ) √
∫ (
) ⁄
Which gives; t =73.8 s
Example (6.3): In the figure shown air is flowing, for which (p=101 kpa and T=5 °C)
and mercury is in the monometer. For (R'=200 mm), calculate the velocity (V)
Sol.:
V=√ ( 0
)
0 000
kg/m
3
0
0
Thus; V=√ ( ) V=205 m/s
a=2
b=
0
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Problems
The problem number listed in the table below refer to the problems in the "text
book", (chapter(8)).
Article No. Related Problems
6.2 1, 2
6.3 3, 4, 5, 6, 8, 10
6.4.1 16, 18, 19, 20, 21, 22,23,24, 25,26, 27, 28, 29, 30, 31, 32, 34, 35, 36,
37, 38
6.4.2 39, 40, 41, 42
6.5 58, 59, 60
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Chapter 7
Closed Conduits Flow Networks
7.1 Introduction
The analysis in this chapter is limited to incompressible fluid ( constant), i.e.,
liquids and gases with very small velocities. Isothermal conditions is assumed to
eliminate thermodynamic effects.
The network considered here consists mainly from (pipes, pumps, fittings, valves,
reservoirs). The analysis will be based on the fundamental principles and equations of
chapter "3" (Continuity and energy equations) and chapter"5" (viscous flow through
pipes, and calculation of the losses for the flow).
7.2 Pipe Friction Formula
In the network analysis, we need a formula that relates the head loss (hf) to the
discharge (Q) of the flow. (hf) here represents the frictional losses if we neglect the minor
losses, and it represents the total losses (frictional +minor) if the minor losses is to be
considered, in this case the length used represents the actual pipe length plus the
equivalent length (Le) for the minor losses. We will list here some of the conventional
formula used for this purpose;
Exponential Pipe –Friction Formula
………………………………. (7.1)
Where;
Head loss per unit length of the pipe
Q = discharge (m3/s)
D = inside diameter (m)
n, m ,R= constants
HazenWilliams Formula
R = 0
n = 1.852 ; m = 4.8704
C= constant depends on pipe roughness
(........................... 7.2)
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Darcy Weisbach Equation
=f
…………………… (7.3)
Equations (7.2) and (7.3) gives;
f = 0
0 0 e 0 ……………………….(7.4)
Conventional Formula
r ……………………………… (7.5)
r=constant depends on pipe
n= exponent (usually 2)
7.3 Concept of Hydraulic and Energy Grade Lines
The Hydraulic Grade Line (HGL) is a line showing the sum of pressure and
potential energies per unit weight (
z), while the energy grade line (EGL) represents
the total energy of the fluid (pressure, potential and kinetic energies per unit weight).
Both lines sloping downward with flow direction due to the losses in energy, except
across a pump, where the energy increases by (hp). Thus;
HGL =
z ………………………………… (7.6)
EGL =
z
…………………….. (7.7)
Hydraulic Gradient =
x(
z) …………………………… (7.8)
Energy Gradient =
x(
z
) …………………. (7.9)
Fig. (7.1)
𝑉
𝑔
𝑉 𝑔
𝑝
ℓ
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Notes:
1. EGL is positioned (v2/2g) above HGL. Thus, if (v=0), as in a large reservoir, they
coincide.
2. Due to losses, EGL always slopes downward in the direction of the flow. The only
exception is when there is a pump:
Fig. (7.2)
3. If there is a turbine, there will be abrupt drop in EGL and HGL. The kinetic energy
can be converted to pressure energy by gradual expansion
If the outlet to a reservoir is an abrupt expansion, all the k.E. is lost, i.e., E.G.L. drops an
amount
at the outlet. If the flow passage changes in diameter, then the slope and the
distance between the EGL and HGL changes.
>
<
Fig. (7.4)
Fig. (7.3)
𝑝
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4. For negative (
), such as in a siphon;
Fig. (7.5)
5. For long pipes,(
) is small compared to (
) and may be neglected. Thus, the
EGL and HGL coincides.
Fig. (7.6)
7.4 Combination of Pipes
7.4.1 Pipes in Series
=……….=
ℓ ∑
………………… (7.10)
E.E.AB Fig. (7.7)
H=
( )
(
)
Equivalent Pipes
Two pipes systems are said to be equivalent when the same head loss produce the
same discharge in both systems: Thus; for the two pipes to be equivalent;
𝑝
𝑝
Є1 Є2 Є3
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Fig. (7.8)
and
Hence;
And;
(
) …………………… (7.11)
Ex.: (D2=250 mm,L2=300 m, f2=0.018) and ( , =0.02). We want to find the
equivalent length of pipe (2) in terms of pipe (1). Using (7.11);
(
)
Thus; a (25.9 m) of (150 mm) pipe is equivalent to (300 m) of (250 mm) pipe for the
assumed conditions.
7.4.2 Pipes in Parallel
Q = Q1 + Q2 + Q3 +…….+ Qn
=(
z ) (
z )
Two types of problems exist;
Type (I)
Given (PA,ZA, PB, ZB), to find Q?
Procedure:
1. (
z ) (
z )=
2. Using the simple pipe problems method, with known, we can find (Q1,Q2,Q3)
from ( ) by using the relations in article (7.2) [hf ]
3. Using C.E.; Q = Q1 + Q2 + Q3
Type (II)
Given (Q) and (D, for each pipe), to find (Q1, Q2, Q3 and , , )
Procedure
1. Assume a discharge ( ) through pipe (1).
2. Calculate ( ) using (
).
............... (7.12)
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Fig. (7.9)
3. Using (
) calculate ( ) and (
).
4. Check C.E., If (
) is equal to the given (Q), then the solution is
over. If not,assume;
You can use ( ) and repeat steps (24) to check.
5. Calculate ( , , )from (
) and check that ( , , )
6. Q1= Q2=
Q3=
7.4.3 Branching Pipes
The following principles have to be applied in solving any
problem involving branching pipes;
1. A the junction (J), C.E. must be satisfied, i.e;
Inflow=Outflow
2. The DarcyWeisbach or any pipe friction formula
should be satisfied for each pipe.
3. There can be only one value of piezometric head (
z) at any point including the
junction (j).
4. The flow will occur in the direction of falling piezometric head in a uniform pipe
length [(
z) (
z) ]
Three types of problems exist for this type of flow. (D, and Pipe friction
formula) are given.
Type "I"
Given Z1, Z2, Z3, Q1 Find Q2,Q3
Procedure:
1. Apply E.E. 1J: (
) (
z)
+ .Here we assume that point (1)
is at the pipe (1) inlet, and we neglect the head of water in the tank; i.e.; and
( ) Thus; (
z) (assumed)
3 25 20 (assumed)
d
Flow
③
①
②
②
③ Z2
Z1
Z3
J
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Fig. (7.10)
2. Since (
z) > then flow is from J ( ) ( )
3. E.E. J (
z)
4. E.E J (
z)
Or; C.E at J: =
Type "II" Given Find:
Type "III" Given Find:
The last two types of problems are solved by trial and error since (
z) cannot be
calculated directly.
Procedure:
1. Assume (
z) ( ) The assumed value could be either larger than z (flow
from(1) (2)&(3)) or smaller than but larger than (flow (1) flow (2) (3))
2. Apply E.E.1 ( )
3. Apply E.E. J (or 2J) =( )
4. Apply E.E J ( )
5. Check C.E at the Junction (J).If ( ), then the solution is over. If not
assume another value of (
z) and repeat the procedure from step (2) until
convergence, the new assume value should satisfy;
If ( > ) . Then assume higher value for (
z) .
If( < ) then assume lower value for (
z)
7.4.4 Pipe Networks
The flow in any network must satisfy the following
three basic conditions;
1. At each junction, the C.E. must be satisfied, i.e.:
Inflow = Out flow
2. The flow in each pipe must satisfy the pipe friction law
AB
C
D
E F
G
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Fig.(7.11)
for flow in a single pipe.
3. The algebraic sum of the head loss around any closed loop must be zero (because we
return to the same point).
“Hardy Cross Method” for pipe network is a systematic method for computation. It
consists from the following steps;
1. Assume the most suitable distribution of flow that satisfy the C.E.
2. Compute ( ) for each pipe ( +ve clockwise and ve counterclockwise).
3. Check condition (3) i.e., ∑ ∑ for each loop.
4. If condition (3) is not satisfied, adjust the flow in each pipe by an amount ( ) for
each loop, i.e.;
and ∑
∑ 0 
∑
∑
0 ………………… (7.13)
The numerator in equ. (7.13) is summed algebraically, while the denominator is
summed arithmetically. Hence, if ( ) it must be subtracted from clockwise
(Qo) and added to counterclockwise ones, and vice versa.
5. The above procedcare is repeated until ( ).
7.5: Pumping Stations
7.5.1: Single Pump in Pipe Line
Consider the single pump in the pipe line system
shown in the figure. The pump must fulfill the following
requirements;
1. Raise the fluid to a static Head (H).
2. Maintaine the flow rate (Q).
3. Overcome the losses in the pipe system.
4. Produce the k. E. with (V2) at the pipe exit.
To calculate the pump head (hp) required for these condition, apply the E. E. from (12),
which gives;
(
∑ )
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(7.12)
Fig. (7.13)
or; by using (V=Q/A), we can show that:
………………. (7.14) “System Curve”
Equ. (7.14) is called the “System Curve” , which gives the head (hp) that must be
supplied to maintain the flow.
To select a pump that fulfills the system requirement equ. (7.14), we have to know
the “pump characteristics”. For each pump, there is a “Pump Characteristics Curves” that
gives the relation between the pump head (hp), pump efficiency ( p or ep) and. The
discharge (Q), i. e. , two curves (hp Q and p ). These characteristics are usually
given in one of the following three forms;
1. Graphs:
2. Tables:
3. Equations:
………………. (7.15)
……………... (7.16)
where (a, b, c, d) are constants (could be + ve or – ve )
Now, if we select a pump whose characteristics are
given (either by curves, tables or equations) and placed it in
the system of Fig. (7.11), the point of intersection of the
system curve (7.14) and the (hp Q) curve of the pump is
called the “Operation Point”.
The designer always try to make the operation point
H
hp
Operation
point
Pump
character
ƺp
ƺp
System
curve
(7.14)
QQop
hp
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Fig. (7.15)
close the point of maximum efficiency.
If the pump characteristics equations (7.15 & 7.16) are used, equs. (7.14) and
(7.15) are solved simultaneously to obtain (Qop), i. e:
Qop = ( )
and then equ. (7.16) is used to obtain ( ).
The table may also be used by assuming a certain value of (Q), calculate (hp) from equ.
(7.14) and compare it with (hp) from table until
convergence.
If there is no point of intersection between the
system curve and the pump characteristics (H > static
head of pump), then either we select other pump with the
larger static head > H), or we can use the principle of
combination of pumps (similar or dis similar ones).
7.5.2 Pumps in Series
For two (or n) pumps connected in series which is used to obtain "High Head " and "Low
Discharge"; see Fig. (7.15);
Q1= Q2 =  =Qn =Qe ………….. (7.17)
hpe =hp1 +hp2++hpn ……….. (7.18)
Ipe= Ip1+Ip2
=
+
Thus,
pe =
 (7.19)
If twin (2), or (n) "identical" pumps are connected in series, then
Q1 = Q2 =  = Qn (identical pump)…….. (7.20)
hp1 = hp2 =  = hpn
Thus; from equs. (7.17) to (7.19); Fig. (7.16)
Fig. (7.14)
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Fig. (7.17)
Qe = Q1 = Q2 =  = Qn ………(7.21)
hpe = nhp1 = nhp2 =  = nphn ………(7.22)
pe = .…….(7.23)
7.5.3 Pumps in Parallel
This connection is used to produce "High Discharge" and "low Head". For two pumps
connected in parallel;
Qe = Q1 + Q2 +  + Qn ……………(7.24)
hpe = hp1 = hp2 =  = hpn ……………..(7.25)
IPe = IP1 + IP2
=
+
Thus;
pe =
……………. (7.26)
If twin (2) or (n) identical pumps are connected in parallel, then;
Q1 = Q2 =  = Qn
(identical pumps) ………… (7.27)
hp1 = hp2 =  = hpn
Thus; from equs. (7.24) to (7.26);
Qe = nQ1 = nQ2 =  = nQn …..…….(7.28)
hpe = hp1 = hp2 =  = hpn ……………..(7.29)
pe = ……….(7. 30)
Note; "High discharge" is required when we want to reduce the time of operation of a
certain flow system (Ex. Filling a tank).
Fig. (7.18)
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Summary of Pumps Connections
If the pump characteristics are given by table or equation form, then the following table
summarizes the rules of connecting (n) identical pumps.
Connection Equation Table
Single Pump hp = a+bQ
2
= cQ+dQ2
hp Q

Increase
max ____
Decrease



Decrease



Increase
(n)Identical
Pumps
Connected In
series
hpe =n(a+bQ2)
= CQe+dQe2
Same
Value
Multiply
Each
Value
by(n)
Same
Value
(n) Identical
Pumps
Connected in
Parallel
hpe=a+b (
)
2
ƺpe= c (
) +d (
)
2
Same
Value
Same
Value
Multiply
Each
Value
by(n)
7.6 Conduits with NonCircular Cross Section
For ducts with noncircular cross sections, the diameter (D) is replaced with
(Dh=4Rh), where);
Dh = Hydraulic Diameter =4 Rh =4
Rh = Hydraulic Radius =
…………… (7.31)
Thus;
hf = f
…………..(7.32)
Re =
…………… (7.33)
……………(7.34) (in Moody Diagram)
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EX:
1. Circular pipe =4
=4
=D
2. Square Duct =4
=a
3. Rectangular Duct =4
( ) =2
Examples
Example (7.1): determine the elevation of hydraulic and energy grade lines at points A,
B, C, D, and E shown in the figure Take (z=3m), and losses due to nozzle as (0.1
z ).
Sol.
E.E. 0E
0+0+23 =0+
+3 +
+0.02
0
0
+10
+ 0.1
 1
C.E.:
(0.15)
2 v =
(0.075)
2 vE
Thus; vE=4 v 2
1 & 2 gives;
= 0.554 and
=16
z
E.E. OA
23= (
+
+z) A + 0.5
Thus; H.G.L. A = (
+z) A =231.5
=23 1.5 *0.554 H.G.L. A=22.17m
E.G.L. A = (
+z+
) A =230.5
=23 0.5 *0.554 E.G.L. A=22.72m
E.E.OB
23= (
+
+z) B + 0.5
+0.02
0
=
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Thus; H.G.L.B = 23(1.5 +0.02
0 )
H.G.L.B=20.4 m
E.G.L. B =H.G.L. B+
= 20.4 +0.55 E.G.L. B=20.95 m
Across the valve, the H.G.L. drops by (10
), or (5.54m). Hence at (c);
H.G.L.C=H.G.L.B5.54 H.G.L.C=14.86m
E.G.L.C=E.G.L.B5.54 E.G.L.C=15.41m
E.E.OD
23= (
+
+z) D +(10.5+0.02
0
0 )
Thus; H.G.L.D=2319.5*0.554 H.G.L.D=12.2m
E.G.L.D=H.G.L.D+
E.G.L.D=12.75m
At point E
H.G.L.E=
+ZE =0+3 H.G.L.E=3m
E.G.L.E=
+
+ ZE =0+16
+3 E.G.L.E=11.86m
Example:(7.2): in the figure shown;
Ki=0.5, L1=300m, D1 =600mm, 1=2mm, L2 =240m, D2=1m, 2=0.3mm, v=3*106
and
H=6m.
Determine the discharge through the system.
Sol.:
Applying E.E. from A to B, using
C.E(v2=v1(
)
2=v1*0.6
2);
6=
[0.5 +f1
00
0 +(10.6
2)
2 +f2
0
0.6
4 +0.6
4]
Thus;
6=
(1.0392+500f1+31.104f2) ……………….. 1
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 165
For (
=0.0033) and (
=0.0003) and the Moody diagram, values of (f) are assumed for
the fully turbulent range;
i.e.; f1=0.026 & f2= 0.015
Thus; from equ. 1 we obtain; v1=2.848 m/s
and v2= 1.025 m/s
Re1=
=
0
0 Re1=569600
Re2=
=
0
0 Re2=341667
Again, from moody diagram, with these (Re1, Re2) and (
,
) we obtain;
f1=0.0265 and f2= 0.0168
From equ. 1 , we obtain v1=2.819 m/s
Thus, Q=A1V1=
D1
2v1 Q=0.797 m
3/s
Ex: (7.3) In the figure shown, L1=900m, D1=300mm, 1=0.3 mm, L2=600m,
D2=200mm, 2=0.03mm, L3=1200m, D3=400mm, 3=0.24mm, =1028kg/m3, v=2.8*10

6m
2/s, pA=560kpa, ZA=30m, ZB=24m.
For a total flow of (340 L/s), determine the flow
through each pipe and the pressure at (B).
Sol.
Assume Q'1 =85 L/s =
=1.2 m/s
Re'1=
=129000
=0.001 Moody Diagram
=0.022
hf'1=
=0.022
00
0
× h
=4.85m
For Pipe 2
h
= h =4.85m =
assuming f'2=0.02 (fully turbulent,
=0.00015), thus.
=1.26 m/s,
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 166
Re'2=
=91400, from Moody Diagram,
=0.019 and =1.291 m/s
Q'2=
D'2 2 Q'2=40.6 L/s
For Pipe 3
hf'3= hf'1 =4.85m = f'3 00
0
with (
=0.0006) assume f'3=0.02, then V'3=1.259 m/s, Re'3=180000, and hence;
f'3=0.02 Q'3=158.2 L/s
The total discharge Q'=Ʃ Q'= Q'1+ Q'2+ Q'3 Q'=283.8 L/s
Hence; Q1=
; Q2=
; Q3=
Thus; Q1= 101.3 L/s Q2 =48.64 L/s Q3 =189.53 L/s
The values of (hf):
v1=
=1.436 m/s Re1=155000 f1=0.021 hf1 =6.62 m
v2=1.548 m/s Re2=109500 f2=0.019 hf2 =6.62m
V3=1.512 m/s Re3=216000 f3=0.019 hf3 =6.64m
Thus, (hf1= hf2= hf3 ) and the solution is over. To find (pB);
E.E.AB
+ ZA =
+ ZB +hf
=
0000
0 +30 24 6.64 = 54.8
Hence PB =54.8*1028*9.8 PB =552.5 kPa
Examle (7.4): In the figure shown; find the dis charges for water at (20˚ c) with the
following pipes data and reservoir elevations:
L1=3000 m D1=1m
=0.0002
L2 = 600m D1= 0.45m
= 0.002
L3 = 1000m D3 =0.6 m
=0.001
Z1=30m Z2=18m Z3=9m
Sol.:
Assume (
+Z)J =23 m
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 167
E.E. 1J 30=23+hf1 hf1 =7=f1
000
f1=0.014 V1=1.75 m/s
Q1=1.38m3/s
E.E. J2 5=f2 00
0
f2=0.024 V2=1.75 m/s Q2=0.278m
3/s
E.E. J3 14=f3 000
0
f3=0.02 V3=2.87 m/s Q3=0.811m
3/s
So that; (Inflow =Q1=1.38 m3/s) is greater than the out flow (Q2+ Q3= 0.278 + 0.811=
1.089) by amount (0.291m3/s).
Assume ((
+Z)J =24.6 m; and similarly);
E.E. 1J 5.4= f1 000
f1=0.015 V1=1.534 m/s Q2=1.205m
3/s
E.E. J2 6.6= f2 00
0
f2=0.024 V2=2.011 m/s Q2=0.32m
3/s
E.E. J3 15.6= f3 000
0
f3=0.02 V3=3.029 m/s Q3=0.856m
3/s
Inflow is still greater by (0.029 m3/s). By extrapolating linearly,
(
+Z)J=24.8m Q1=1.183m
3/s Q2=0.325m
3/s Q3=0.856m
3/s
Example (7.5): Calculate the discharge through each pipe of the network shown in the
figure. Take (n=2).
Sol.:
First Trail
Loop 1 Loop 2
r Q0n n rQ0
n1 r Q0
n n rQ0
n1
1*602=3600 2*1*60=120
3*52=75 2*3*5=30
2*402=3200 2*2*40=160
Ʃ r Q0n=475 Ʃǀn rQ0
n1 ǀ=310
∆Q= 
0 ≈ 1.5
2*302 =1800 2*2*30=120
3*52=75 2*3*5=30
1*452=2025 2*1*45=90
ƩrQ0n =300 Ʃ n rQ0
n1ǀ=240
∆Q=  00
0 ≈ 1
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 168
Trial 2 Q=Q0+∆Q
2.5=5+(1.5)1
Thus ∆Q≈0 for all loops, and hence the final distribution will be;
Example (7.6): A pump characteristics are given
by; (hp =31.12000Q2) and (ƺp =190000Q
2+8590Q),
where (hp) is in (m) and (Q) in (m3/s) and (ƺp) is the
efficiency (%).
a. Find the characteristics of three identical pumps
connected in parallel
b. For the piping system shown in the figure,
estimate the difference in operational cost between one pump and three pumps in
parallel to supply (1000m3) of water to the upper reservoir if the cost of (1KWh) is
(65 fils). Neglect minor losses
c. Repeat (a) and (b) for three identical pumps in series.
Sol.:
(a) Qe=3Q1=3Q2=3Q3=3Q
hpe=hp1=hp2=hp3 hpe=3112000 (
)2
ƺpe = ƺp1= ƺp2= ƺp3 ƺpe=190000(
)2
+8590(
)
Loop 1 Loop2
r Q0n n rQ0
n1 r Q0
n n rQ0
n1
1*58.52=3422.25 117
3*2.52=18.75 15
2*41.52=3444.5 166
Ʃ r Q0n =3.5 Ʃǀn rQ0
n1 ǀ=298
∆Q= 
=0.012
2*312=1922 124
3*2.52=19 15
1*442=1936 88
Ʃ r Q0n =33 Ʃǀ n rQ0
n1 ǀ=227
∆Q= 
=0.145
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 169
(b) E.E hp =15+hf=15+f
Or ; hp =15+65352Q2 (1)
One pump
3112000Q2=15+65352Q
2
Q=0.01438m
3/s
hp=3112000(0.01438)2 hp=28.51m
p =190000(0.01438)2+8590(0.01438) p=84.23%
Ip=
Ip=4774.5W
t =
=
000
0 0 t=695410.3 s
no. of KWh= Ip*t=
000 * 0
00 no. of KWh=922.3
cost. =922.3*0.065 cost =59.95 ID
Three Pumps
Qe=3Q1=3*0.01438 Qe=0.04314m3/s
hpe =hp1=hp2=hp3 hpe =28.51m
pe = p1 = p2= p3 pe =84.23%
t=
=
000
0 0 t=231803.4 s
Ip=
Ip=14324.5 W
no.of KWh= Ip*t no. of KWh=922.3
cost. =922.3*0.065 cost =59.95 ID
Thus, there is no difference in cost. Only the time will be reduced
(c) Qe=Q1=Q2=Q3=Q
hpe=3Q1=3Q2=3Q3 hpe=9336000 Q2
ƺpe = ƺp1= ƺp2= ƺp3 ƺpe=190000Q2+8590
One pump; the same as in (b)
Three Pumps in Series
Qe=Q1=Q2=Q3=Q Qe=0.01438 m3/s
hpe =3 hp =3*28.51 hpe =85.53
ƺpe = ƺp1= ƺp2= ƺp3 ƺpe=84.23%
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 170
Ip=
Ip=14324.5 W
t=
=
000
0 0 t=695410.3 s
no. of KWh= Ip*t no. of KWh=2767
Cost. =2767*0.065 cost. =179.86 ID
Cost is higher than for single pump.
Example(7.7): Determine the head loss, in millimeters of water, required for flow of
(300 m3/min) of air at (20˚c) and (100 kpa) through a rectangular galvanizediron section
(700 mm) wide, (350mm) high, and (70m) long.
Sol.:
Rh=
=
0 0
(0 0 ) Rh=0.117 m
Dh=2 Rh Dh=0.468m
=
=
0 000
0
= 0.00032
V=
=
00 0
0 0 v=20.41 m/s
=
=
00000
=1.189 kg/m
3
From Fig. (c.1); =2.2*105
Pa. s
Re=
=
0 0
0 Re=516200
From Moody Diagram; f=0.0165
Thus;
hf= f
= 0.0165 0
0 0
hf=52.42 m air
ɣ air = air g =1.189*9.8 ɣ air =11.66 N/m3
Thus;
hf (mH2O)= hf *
= 52.42 *
0 O O
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 171
Problems
The problems number listed in the table below refer to the problems in the
"text book", chapter (10)
Article No. Related Problems
7.3 1,2,3,5,6,7,8
7.4 10,11,15,17,19,28,31,32,38,40,41
7.5 20,22,23,24,25,26,27,33,34,35,36,37,39
7.6 44, 45,46,47
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 172
Appendix –A
COURSE FOLIO
Course Title – Symbol: Fluid Mechanics / I – ME202
Instructor:
Prof. Dr. Ihsan Y. Hussain
Professor of Mechanical Engineering / ThermoFluids
Mech. Engr. Dept.
College of Engineering
University of Baghdad
Tel: +009647901781035
Email: [email protected] ; [email protected] :
Teaching Assistant:
Dr. Ammar A. Farhan
Lecturer of Mechanical Engineering / ThermoFluids
Mech. Engr. Dept.
College of Engineering
University of Baghdad
Tel: +009647903566684
Email: [email protected]
Course Description (or Catalog Data)
ME202  Fluid Mechanics / I
This course introduces the description of phenomena associated with fluid flow.
Topics covered: physical properties of fluids; fluid statics; principles of conservation of
mass, energy and momentum; control volume technique; Bernoulli equation; dimensional
analysis and similitude; viscous flow in pipes and channels; laminar and turbulent flow;
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 173
boundary layer theory; drag and lift; Moody diagram; pipe problems; flow and fluid
measurements; analysis of pipes and pumps networks. Physical understanding of fluid
flows and applications to practical problems will be stressed. The course is designed to
provide a background to higher level courses involving fluid flow and heat transfer. The
course is taught through 5 hrs per week, 3 theories, 1 tutorial, and 1 experimental.
Prerequisites: ME101 & ME102 Courses
Goals / Objectives
1. Introduce basic definitions and introductory concepts of fluid mechanics.
2. Introduce the description of pressure distribution in a static fluid and its effects on
submerged surfaces and bodies.
3. Introduce the description of phenomena associated with fluid flow phenomena.
4. Explain and derive the conservation laws that govern fluid motion ( continuity,
energy, and momentum equations).
5. Introduce the principles of “Dimensional Analysis” and “Similitude” and their
application to fluid mechanics problems.
6. Introduce the principles of viscous flow, boundary layer, drag and lift, primary and
secondary losses in pipe flow.
7. Enable the student to analyze and design pipes network and pumps connection.
8. Enable the student to measure the fluid properties and flow parameters, and to
design and conduct experiments of fluid mechanics.
9. Provide a strong physical and analytical understanding of fluid
flows in order to function in the capacity of mechanical engineer in
an engineering company dealing with fluid machinery.
10. Provide a background to higher level courses involving fluid flow
and head transfer.
Student Learning Outcomes
At the end of the class, the student will be able to:
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 174
a. Define Fluids and Fluid Mechanics and distinguish between
incompressible and compressible fluids, and understand and define
the basic fluid properties; especially density and viscosity, and
apply Newton’s law of viscosity.
b. Calculate; the pressure in static fluid, hydrostatic forces on
submerged surfaces, buoyancy forces, stability of submerged and
floating bodies, Metacenter, and forces on accelerated fluids.
c. Be familiar with continuity, energy, and momentum equations, and
their applications to fluid flow problems.
d. Understand and apply the principles of dimensional analysis and
similitude to fluid mechanics problems.
e. Formulate and solve incompressible laminar flows for simple
parallel flows in Cartesian and polar coordinates.
f. Analyze boundary layer flows over flat plate.
g. Estimate drag and lift forces in laminar and turbulent flows for
different immersed bodies.
h. Calculate frictional losses in pipe problems for both laminar and
turbulent flows, by using Moody Diagram.
i. Calculate secondary ( minor ) losses for various pipes fittings and
connections.
j. Know how to measure flow properties ( pressure, velocity,
discharge ) and fluid properties ( density and viscosity ).
k. Be able to analyze and design pipes network and connection, and
pumping stations and connection.
l. Be able to apply modern knowledge and to apply mathematics,
science, engineering and technology to fluid mechanics problems
and applications.
m. Design and conduct experiments of fluid mechanics, as well as
analyze, interpret data and apply the experimental results for the
services.
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 175
n. Work in groups and function on multidisciplinary teams.
o. Identify, formulate and solve engineering fluid mechanics
problems.
p. Understand professional, social and ethical responsibilities.
q. Communicate effectively.
r. Use the techniques, skills, and modern engineering tools necessary
for engineering practice in fluid mechanics applications.
Course Schedule
Hours / Week Theoretical Content Week
Exp. Tut. Theo.
1 1 3 Introductory Concepts To Fluid
Mechanics 1
1 1 3 Introductory Concepts To Fluid
Mechanics 2
1 1 3 Fluid Statics : Pressure Distribution In
Static Fluids 3
1 1 3 Pressure Measurements 4
1 1 3 Forces On Immersed Surfaces 5
1 1 3 Forces On Immersed Surfaces 6
1 1 3 Buoyancy And Floatation 7
1 1 3 Buoyancy And Floatation 8
1 1 3 Buoyancy And Floatation 9
1 1 3 Accelerated Fluid And Relative
Motion 01
1 1 3 Introduction To Fluid Motion 00
1 1 3 Continuity Equation 02
1 1 3 Energy Equation 03
1 1 3 Momentum Equation 04
1 1 3 Momentum Equation 05
1 1 3 Dimensional Analysis And Similitude 06
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 176
1 1 3 Dimensional Analysis And Similitude 07
1 1 3 Dimensional Analysis And Similitude 08
1 1 3 Laminar Viscous Flow Between Parallel
Plates 09
1 1 3 Laminar Viscous Flow Through Circular
Tubes 21
1 1 3 Boundary Layer Theory, Drag & Lift 20
1 1 3 Losses In Pipes : Moody Diagram 22
1 1 3 Losses In Pipes : Moody Diagram 23
1 1 3 Losses In Pipes : Moody Diagram 24
1 1 3 Measurements Of Fluid Flow 25
1 1 3 Measurements Of Fluid Flow 26
1 1 3 Measurements Of Fluid Flow 27
1 1 3 Analysis Of Piping And Pumping
Networks 28
1 1 3 Analysis Of Piping And Pumping
Networks 29
1 1 3 Analysis Of Piping And Pumping
Networks 31
Textbook and References
1. “Fluid Mechanics”; by Victor L. Streeter and E. Benjamin Wylie,
First SI Metric Edition, M G. GNW Hill , 1988.
2. “Fundamental of Fluid Mechanics”; by Bruce E. Munson,
Theodore H. Okiishi, and Wade W. Huesch, Benjamin
Wylie, Sixth Edition, 2009
3. “Fluid Mechanics : Fundamentals and Applications”; by
Yunus A. Çengel and John M. Cimbala, M G. GNW Hill
Higher Education, 2006
4. “Introductory Fluid Mechanics” ; by Joseph Katz,
Cambridge University Press, 2010
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 177
5. “Elementary Fluid Mechanics”, by John K. Vennard and
Robert L. Streat, 5th
ed., John Wiley and Sons, 1976.
6. “Engineering Fluid Mechanics by John A. Robert and
Clayton T. Crow, 2nd
ed., Houghton Mifflin Coo, 1988.
Grading Units
Annual Semester System Academic
System
Final
Examination
Laboratory
Work Quest
Course
Assessment for
Annual System
(011)%
70% \
30 % ( Tests, , Quizzes, and
Extracurricular Activities )
1 The laboratory experiments are included in the
general course (Laboratories / 2).
2 Two  Three tests are made; and not Less than (20
25) Quizzes are usually made.
Additional
Information
Typical Grading
Excellent 90100%
Very Good 8089%
Good 7079%
Fair 6069%
Pass 5059%
Poor <50%
Grading Policy
1. Q uizzes:
 There will be a ( 20 – 25 ) closed books and notes quizzes during
the academic year.
 The quizzes will count 20% of the total course grade.
2. Tests, 23 Nos. and will count 10% of the total course grade.
3. Extracurricular Activities, this is optional and will count extra marks
( 1 – 5 % ) for the student, depending on the type of activity.
4. Final Exam:
√
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 178
 The final exam will be comprehensive, closed books and notes,
and will take place on
(Saturday8th
) , ( January ), 2013 from 9:00 AM  12:00 PM in
rooms ( M12 + M13 )
 The final exam will count 70% of the total course grade
Assessment Plan
1. Reinforcement is done through tests, quizzes, extracurricular
activities and student engagement during lectures as shown in the
table below.
Course Outcomes Strategies/Actions Assessment Methods
a) Define Fluids and Fluid
Mechanics and distinguish
between incompressible
and compressible fluids,
and understand and define
the basic fluid properties;
especially density and
viscosity, and apply
Newton’s law of viscosity.
Lecture plan and inclass
activities: each class will
commence with a
summary of the
previous lecture,
questions will be asked
and the responses will be
used to evaluate the
students’ understanding
of the topics covered.
Inclass questions and
discussion
Quizzes
Lab. Experiments
b) Calculate; the pressure in
static fluid, hydrostatic
forces on submerged
surfaces, buoyancy
forces, stability of
submerged and floating
bodies, Metacenter, and
forces on accelerated
fluids.
Lecture plan and inclass
activities: each class will
commence with a
summary of the
previous lecture,
questions will be asked
and the responses will be
used to evaluate the
students’ understanding
of the topics covered.
Inclass questions and
discussion
Quizzes
Lab. Experiments
c) Be familiar with
continuity, energy, and
momentum equations, and
their applications to fluid
flow problems.
Lecture plan and inclass
activities: each class will
commence with a
summary of the
previous lecture,
questions will be asked
and the responses will be
used to evaluate the
students’ understanding
of the topics covered.
Inclass questions and
discussion
Quizzes
Lab. Experiments
d) Understand and apply
the principles of
Lecture plan and inclass
activities: each class will
Inclass questions and
discussion
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 179
dimensional analysis
and similitude to fluid
mechanics problems.
commence with a
summary of the
previous lecture,
questions will be asked
and the responses will be
used to evaluate the
students’ understanding
of the topics covered.
Quizzes
Lab. Experiments
e) Formulate and solve
incompressible laminar
flows for simple parallel
flows in Cartesian and
polar coordinates.
Lecture plan and inclass
activities: each class will
commence with a
summary of the
previous lecture,
questions will be asked
and the responses will be
used to evaluate the
students’ understanding
of the topics covered.
Inclass questions and
discussion
Quizzes
Lab. Experiments
f) Analyze boundary layer
flows over flat plate.
Lecture plan and inclass
activities: each class will
commence with a
summary of the
previous lecture,
questions will be asked
and the responses will be
used to evaluate the
students’ understanding
of the topics covered.
Inclass questions and
discussion
Quizzes
Lab. Experiments
g) Estimate drag and lift
forces in laminar and
turbulent flows for
different immersed
bodies.
Lecture plan and inclass
activities: each class will
commence with a
summary of the
previous lecture,
questions will be asked
and the responses will be
used to evaluate the
students’ understanding
of the topics covered.
Inclass questions and
discussion
Quizzes
Lab. Experiments
h) Calculate frictional
losses in pipe problems
for both laminar and
turbulent flows, by
using Moody Diagram.
Lecture plan and inclass
activities: each class will
commence with a
summary of the
previous lecture,
questions will be asked
and the responses will be
used to evaluate the
students’ understanding
of the topics covered.
Inclass questions and
discussion
Quizzes
Lab. Experiments
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 180
i) Calculate secondary (
minor ) losses for
various pipes fittings
and connections .
Lecture plan and inclass
activities: each class will
commence with a
summary of the
previous lecture,
questions will be asked
and the responses will be
used to evaluate the
students’ understanding
of the topics covered.
Inclass questions and
discusion
Quizzes
Lab. Experiments
j) Know how to measure
flow properties (
pressure, velocity,
discharge ) and fluid
properties ( density and
viscosity ).
Lecture plan and inclass
activities: each class will
commence with a
summary of the
previous lecture,
questions will be asked
and the responses will be
used to evaluate the
students’ understanding
of the topics covered.
Inclass questions and
discussion
Quizzes
Lab. Experiments
k) Be able to analyze and
design pipes network
and connection, and
pumping stations and
connection.
Lecture plan and inclass
activities: each class
will commence with a
summary of the
previous lecture,
questions will be asked
and the responses will be
used to evaluate the
students’ understanding
of the topics covered.
Inclass questions and
discussion
Quizzes
Lab. Experiments
l) Be able to apply modern
knowledge and to apply
mathematics, science,
engineering and
technology to fluid
mechanics problems and
applications.
Lectures
Experiments
Tests and Exams
Extracurricular
activities
Inclass questions and
discussion
Quizzes
Lab. Experiments
Extracurricular
activities
m) Design and conduct
experiments of fluid
mechanics, as well as
analyze, interpret data
and apply the
experimental results for
the services.
Conducting
experiments in Fluid
Mechanics Lab. and
technical writing of
reports including
discussions and
conclusions
Inclass questions and
discussion
Quizzes
Lab. Experiments
n) Work in groups and
function on multi
disciplinary teams.
Conducting
experiments in
groups
Tutorial hour work
Extracurricular
Monitoring, Guiding,
Evaluation, Quizzes …..
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 181
activities
o) Identify, formulate and
solve engineering fluid
mechanics problems.
Homework , Tutorial
hour, Solved Examples
…..
Inclass questions and
discussion
Quizzes
Tests
p) Understand professional,
social and ethical
responsibilities.
In and OutClass oral
conservations,
lectures….
Discussions and
conservations
Extracurricular
activities
q) Communicate
effectively.
In and OutClass oral
conservations,
lectures….
Discussions and
conservations
Extracurricular
activities
r) Use the techniques,
skills, and modern
engineering tools
necessary for
engineering practice in
fluid mechanics
applications.
In and OutClass oral
conservations,
lectures….
Discussions and
conservations
Extracurricular
activities
s) Be able to analyze and
design pipes network
and connection, and
pumping stations and
connection.
Lecture plan and inclass
activities: each class
will commence with a
summary of the
previous lecture,
questions will be asked
and the responses will be
used to evaluate the
students’ understanding
of the topics covered.
Inclass questions and
discussion
Quizzes
Lab. Experiments
t) Be able to apply modern
knowledge and to apply
mathematics, science,
engineering and
technology to fluid
mechanics problems and
applications.
Lectures
Experiments
Tests and Exams
Extracurricular
activities
Inclass questions and
discussion
Quizzes
Lab. Experiments
Extracurricular
activities
u) Design and conduct
experiments of fluid
mechanics, as well as
analyze, interpret data
and apply the
experimental results for
the services.
Conducting
experiments in Fluid
Mechanics Lab. and
technical writing of
reports including
discussions and
conclusions
Inclass questions and
discussion
Quizzes
Lab. Experiments
v) Work in groups and
function on multi
disciplinary teams.
Conducting
experiments in
groups
Tutorial hour work
Monitoring, Guiding,
Evaluation, Quizzes …..
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 182
Extracurricular
activities
w) Identify, formulate and
solve engineering fluid
mechanics problems.
Homework , Tutorial
hour, Solved Examples
…..
Inclass questions and
discussion
Quizzes
Tests
x) Understand professional,
social and ethical
responsibilities.
In and OutClass oral
conservations,
lectures….
Discussions and
conservations
Extracurricular
activities
y) Communicate
effectively.
In and OutClass oral
conservations,
lectures….
Discussions and
conservations
Extracurricular
activities
z) Use the techniques,
skills, and modern
engineering tools
necessary for
engineering practice in
fluid mechanics
applications.
In and OutClass oral
conservations,
lectures….
Discussions and
conservations
Extracurricular
activities
2. Lists the responses obtained from student survey conducted at the end of
academic year. A student’s opinion questionnaire was made for a selected
specimen of the students. The results of this questionnaire are shown in
Table ( 1 ) and figure ( 1 ) for the students opinion about curriculum , and in
Table ( 2 ) and figure ( 2 ) for the students opinion about faculty member.
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 183
Table ( 1 ): Students Opinion Questionnaire about Curriculum
Academic Year 2010 – 2011
Code No. & Curriculum Name: : ME 202 FLUID MECHANICS / I
Year: 2nd
Year
Faculty Member's Name: Prof. Dr. Ihsan Y. Hussain
Dear Students: For the development of the educational process at the university, we
hope to express your opinion by answering accurately with mark √ in the place
which reflects your opinion taking into consideration the accuracy and objectivity.
Score 1 2 3 4 5
No. Question Strongly
Agree Agree
I
Don’t
Know
Disagree
I Don’t
Agree At
All
1 Overall, this Curriculum subject is
good and useful 9 7 0 0 0
2 Lecture time is sufficient to cover
the contents of the article 7 6 0 3 0
3 The content of article commensurate
with the objective of Curriculum 6 9 1 0 0
4 Subject content is an interdependent
information 5 7 1 3 0
5 Textbooks and references are
available and meaningful 3 7 1 5 0
6 available of References helpful for
stimulate and thinking 2 8 2 1 3
7 The book is free of grammatical
errors Printing 3 6 5 2 0
8 Contents of the book are of outdated
information 2 6 7 0 1
9 The book contains a variety of
examples and exercises 7 7 1 1 0
10 The evaluation of the subject system
is appropriate (test method) 6 7 1 0 2
11 Exams reflect the content of the
subject 6 6 1 3 0
12 Number of exams be exhaustive of
the content subject 6 8 2 0 0
13 Examinations and assignments
helped to absorb the subject 5 8 0 0 3
14 Examinations and exercises are in
line with the objectives of the
subject
5 8 1 2 0
15 Examinations and exercises help to
think of more conservation 5 9 0 2 0
16 Number of exams and the their
recurrence are appropriate 6 6 2 1 1
17 The case of equipped lecture halls
satisfactory 2 3 0 5 6
18 Capabilities and laboratories are
appropriate and effective 1 1 6 8
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 184
Table ( 2 ): Students Opinion Questionnaire about Faculty Member
Academic Year 2010  2011
Code No. & Curriculum Name: ME 202  FLUID MECHANICS (1)
Year: 2nd
Year Faculty Member's Name: Dr. Ihsan Y. Hussain
Is the plan of teaching the subject was distributed from the beginning of the year? Yes
No I don’t know
Is the faculty member is committed to the specific office hours of the subject? Yes No
I don’t know
If the answer is (No) explained that___________
Dear Students: For the development of the educational process at the university we hope to
express your opinion by answering accurately with mark √ in the place which reflects your
opinion taking into consideration the accuracy and objectivity.
Score 1 2 3 4 5
No. Question Strongly
Agree Agree
I
Don’t
Know
Disagree
I Don’t
Agree At
All
1 Has the ability to communicate
scientific material in a smooth and
easy manner
2 11 1 1 1
2 Keep to use the tools and techniques
of modern education 1 2 2 9 2
3 Illustrates the theoretical aspects in
the subject with examples from the
reality
3 11 0 2 0
4 Gives the scientific material in a
manner covering the time of the
lecture
3 8 1 4 0
5 Committed to the dates of lectures 12 4 0 0 0
6 Improve in the management ranks
and give equal opportunities to
students in dialogue and discussion
1 12 1 1 1
7 Motivates students and encourages
them to think and research 0 10 1 4 1
8 Respects the different views of the
students 1 13 1 1 0
9 Through selflearning encourages
students to search for what is new
and modern
1 7 2 5 1
10 Accept criticism and suggestions
with an open mind 0 7 7 1 1
11 Be objective and fair in his / her
evaluation of students 2 10 3 1 0
12 Uses a variety of methods to assess
the performance of students (such as
reports, research, and quizzes),
4 4 1 7 1
13 Follow up activities and duties to
put the evaluation weights 0 2 5 9 0
14 Has the ability to discuss all issues
of the subject 10 4 1 0 1
15 Working to increase the knowledge
of the outcome requested 2 10 1 2 1
3
13
12 1
3
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 185
Figure (1): Students Opinion Questionnaire about Fluid Mechanics (1)
Curriculum
Figure (2): Students Opinion Questionnaire about Fluid Mechanics (1)
Instructor
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
% o
f A
nsw
er
for
Each
Se
rie
s
Question No.
Students Opinion  Curricula
Series1 Series2 Series3 Series4 Series5
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
% o
f A
nsw
er
for
Each
Se
rie
s
Question No.
Students Opinion  Faculty Member
Series1 Series2 Series3 Series4 Series5
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 186
3. Students rating performance: Table ( 3 ) shows the results of quizzes and tests related
to some of the course outcomes.
Table ( 3 ): Students Rating Related to some of the Course LOs
Mark ( 10 % )
Theoretical Content LO %
Success St. Dev. Avg.
48.8 % 4.7 Introductory Concepts To Fluid
Mechanics
a
58.5 % 5.1 Introductory Concepts To Fluid
Mechanics
33.3 % 3.95 Fluid Statics : Pressure Distribution In
Static Fluids & Pressure Measurements
b 31.1 % 3.8 Forces On Immersed Surfaces
13 % 2.1 Buoyancy And Floatation
17.5 % 3.1 Accelerated Fluid And Relative
Motion
84.4 % 6.7 Continuity Equation
c 29.8 % 3.05 Energy Equation
4 % 1 Momentum Equation
Dimensional Analysis And Similitude
d Dimensional Analysis And Similitude
Dimensional Analysis And Similitude
30.1 % 3.15 Test / I a, b,
c, d
68.9 % 5 Laminar Viscous Flow Between Parallel
Plates
e
40 % 3.9 Laminar Viscous Flow Through Circular
Tubes
52.9 % 5 Boundary Layer Theory f
58.5 % 5.35 Drag & Lift g
Losses In Pipes : Moody Diagram h &
i Losses In Pipes : Moody Diagram
Incompressible Fluid Mechanics 2017
Prof. Dr. Ihsan Y. Hussain / Mech. Engr. Dept.  College of Engr. – University of Baghdad
Page 187
Losses In Pipes : Moody Diagram
Measurements Of Fluid Flow
j Measurements Of Fluid Flow
Measurements Of Fluid Flow
Analysis Of Piping And Pumping
Networks
k Analysis Of Piping And Pumping
Networks
Analysis Of Piping And Pumping
Networks
Results of Tables 1, 2, and 3 are used in Table 4 to assess changes/improvements from
the previous evaluation as well as changes/improvements that will be made to the course
to improve student achievement of the learning outcomes.
Table ( 4 ): Changes/Improvements Term
Assessment of
changes/improvements
made in previous term
Good improvement in
experimental part
Good response from students
for the extracurricular
activities
Good response for the case of
the subject submitted to the
students, which includes
lectures and modern
references
Changes/improvements
that will be made next
time the subject is
offered
Use the tools and techniques of
modern education
Use of Interactive Class
Using scientific films and videos in teaching the subject
Increase the number of extracurricular activities
Decrease the number of quizzes
Improve Laboratory equipment